🧠:<think>Alright, so I've got this problem from Dr. Elena Voss, and I need to figure out the orbital period of a binary star system and then determine the change in the orbital period of a space station due to a gravitational perturbation. Hmm, okay, let's take it step by step.Starting with the first part: Orbital Mechanics. We have two stars, M1 and M2, orbiting their common center of mass in circular orbits. The distance between them is d. I need to derive the expression for the orbital period T in terms of M1, M2, and d.I remember that in a two-body system, both objects orbit their common center of mass. The gravitational force between them provides the necessary centripetal force for their circular orbits. So, the gravitational force F_gravity is equal to the centripetal force F_centripetal.The gravitational force between the two stars is given by Newton's law of universal gravitation:F_gravity = G * (M1 * M2) / d²Where G is the gravitational constant.Now, each star is moving in a circular orbit around the center of mass. Let's denote the distance from M1 to the center of mass as r1, and from M2 as r2. Since they orbit the center of mass, we know that M1 * r1 = M2 * r2. Also, the total distance between the two stars is d = r1 + r2.So, we can express r1 and r2 in terms of M1, M2, and d. Let me solve for r1 and r2.From M1 * r1 = M2 * r2, we get r2 = (M1 / M2) * r1.Substituting into d = r1 + r2:d = r1 + (M1 / M2) * r1 = r1 (1 + M1 / M2) = r1 ( (M2 + M1) / M2 )Therefore, r1 = d * (M2 / (M1 + M2))Similarly, r2 = d * (M1 / (M1 + M2))Okay, so now each star is moving in a circular orbit with radius r1 and r2 respectively. The centripetal force required for each star's orbit is provided by the gravitational force.So, for M1:F_centripetal = M1 * (v1² / r1) = (M1 * v1²) / r1Similarly, for M2:F_centripetal = M2 * (v2² / r2) = (M2 * v2²) / r2But since the gravitational force is the same for both, we can set F_gravity equal to both expressions.So, G * (M1 * M2) / d² = (M1 * v1²) / r1 = (M2 * v2²) / r2Let me pick one of these to work with. Let's take the first one:G * (M1 * M2) / d² = (M1 * v1²) / r1We can cancel M1 from both sides:G * M2 / d² = v1² / r1Similarly, for the second equation:G * M1 / d² = v2² / r2But since both stars have the same orbital period T, their velocities are related to their orbital radii and the period.The orbital velocity v is given by v = 2πr / T.So, for M1: v1 = 2π r1 / TFor M2: v2 = 2π r2 / TLet me substitute v1 into the first equation.G * M2 / d² = ( (2π r1 / T )² ) / r1Simplify the right-hand side:(4π² r1² / T² ) / r1 = 4π² r1 / T²So, G * M2 / d² = 4π² r1 / T²We can solve for T²:T² = (4π² r1 d²) / (G M2)But we know that r1 = d * (M2 / (M1 + M2)), from earlier.Substituting r1:T² = (4π² * (d * M2 / (M1 + M2)) * d² ) / (G M2 )Simplify numerator:4π² * d³ * M2 / (M1 + M2)Denominator: G M2So, T² = (4π² d³ M2) / (G M2 (M1 + M2)) )The M2 terms cancel:T² = (4π² d³) / (G (M1 + M2))Therefore, T = sqrt( (4π² d³) / (G (M1 + M2)) )Which can be written as:T = 2π * sqrt( d³ / (G (M1 + M2)) )Hmm, that seems familiar. It's similar to Kepler's third law, which states that the square of the orbital period is proportional to the cube of the semi-major axis, with the constant involving the sum of the masses.So, that seems correct. Let me just recap:1. Calculated gravitational force between the two stars.2. Expressed the centripetal force for each star.3. Related the velocities to the orbital period.4. Substituted and solved for T, leading to the expression involving the sum of the masses and the distance cubed.Okay, that seems solid.Now, moving on to the second part: Gravitational Perturbation.We have a space station orbiting a planet of mass M_p and radius R_p. The space station is at an altitude h above the planet's surface. So, the orbital radius of the space station is R = R_p + h.We need to determine the change in the orbital period due to a nearby moon of mass M_m and distance D from the planet. The perturbation is small, so we can treat it as a first-order correction.So, the unperturbed orbital period is the period of the space station around the planet, and the perturbation is due to the moon's gravitational pull. Since the perturbation is small, we can approximate the change in period using a first-order term.First, let me recall the unperturbed orbital period of the space station.Using Kepler's third law, the period T0 is:T0 = 2π * sqrt( R³ / (G M_p) )Where R = R_p + h.Now, the perturbation due to the moon. The moon is at a distance D from the planet, so the distance between the moon and the space station is variable, but since the perturbation is small, perhaps we can approximate it as a constant perturbation or average it over the orbit.But wait, the moon is orbiting the planet as well, right? So, the distance between the space station and the moon would vary depending on their relative positions. However, since the perturbation is small, maybe we can model it as a tidal force or consider the average effect.Alternatively, perhaps we can consider the gravitational acceleration due to the moon at the location of the space station and then find the resulting change in the orbital velocity or period.Let me think.The gravitational acceleration from the moon at the space station's location would be:a_moon = G M_m / r²Where r is the distance between the moon and the space station.But the moon is orbiting the planet at a distance D, so depending on the relative positions, r can vary between D - R and D + R, where R is the space station's orbital radius.However, if the moon's orbit is much larger than the space station's orbit, then r ≈ D, so the perturbation is approximately G M_m / D².But if the moon is close, then the perturbation varies significantly. Since the problem says the perturbation is small, perhaps we can approximate the perturbation as a constant acceleration at the space station's location.Alternatively, perhaps we can model the perturbation as a tidal force, which is the difference in gravitational acceleration across the space station's orbit.But maybe a simpler approach is to consider the additional gravitational acceleration from the moon and find how it affects the orbital period.Wait, but the orbital period is determined by the balance between gravitational force and centripetal acceleration. If there's an additional gravitational force from the moon, it would perturb the orbit, causing a change in the orbital period.But since the perturbation is small, we can treat it as a first-order correction.Let me denote the unperturbed period as T0 and the perturbed period as T = T0 + ΔT, where ΔT is the small change.Alternatively, perhaps it's better to consider the change in the effective gravitational parameter.Wait, in the unperturbed case, the space station orbits the planet with gravitational parameter μ = G M_p.If we consider the moon's gravity as a perturbation, it effectively adds a small gravitational acceleration, which can be considered as an additional term in the gravitational force.But since the moon is at a distance D, the gravitational acceleration at the space station's location due to the moon is approximately G M_m / D², assuming D is much larger than R (the space station's orbital radius). If D is not much larger, then we might have to consider the difference in acceleration across the space station's orbit, leading to tidal forces.But the problem states that the perturbation is small, so perhaps we can model it as a uniform acceleration, or maybe as a perturbation to the effective gravitational parameter.Wait, another approach: The orbital period is determined by the balance of forces. The perturbing force from the moon would cause a small change in the net gravitational force, which would lead to a small change in the orbital velocity, and hence a change in the orbital period.Let me denote the unperturbed gravitational acceleration as a0 = G M_p / R².The perturbing acceleration from the moon is a_moon = G M_m / D².Since the perturbation is small, a_moon << a0.Therefore, the total acceleration is approximately a0 + a_moon.But wait, actually, the moon's gravitational acceleration is not in the same direction as the planet's. Depending on the position, it could be in any direction. However, since we're looking for a first-order correction, perhaps we can consider the average effect or the maximum effect.But maybe a better approach is to consider the perturbation as a small change in the effective gravitational parameter.Wait, let's think in terms of the equation for orbital period.In the unperturbed case:T0 = 2π sqrt(R³ / (G M_p))If we consider the perturbation, the effective gravitational parameter becomes G(M_p + δM), where δM is a small correction due to the moon's gravity.But that might not be directly applicable because the moon's gravity is not central with respect to the space station's orbit.Alternatively, perhaps we can model the perturbation as an additional gravitational force, which would cause a small change in the orbital radius or period.Wait, another idea: The perturbing force from the moon can be considered as a small perturbation to the gravitational potential. The change in the orbital period can be found by considering the change in the effective potential.Alternatively, perhaps we can use the concept of the Roche limit or tidal forces, but that might be more complicated.Wait, maybe a simpler approach is to consider the perturbation as a small additional acceleration, which would cause a small change in the orbital velocity, and hence a small change in the orbital period.Let me denote the unperturbed velocity as v0 = sqrt(G M_p / R).The perturbing acceleration a_moon = G M_m / D².Assuming that the perturbing acceleration is in the same direction as the unperturbed acceleration (which might not be the case, but for a first-order approximation, perhaps we can consider it as such), the total acceleration would be a0 + a_moon.But actually, the perturbing acceleration is not necessarily in the same direction. It could be in any direction depending on the moon's position. However, since we're looking for a first-order correction, maybe we can consider the average effect over the orbit.Alternatively, perhaps we can consider the perturbation as a small change in the effective gravitational parameter.Wait, let's think about the equation for the orbital period again.T = 2π sqrt(R³ / (G M_eff))Where M_eff is the effective mass. If the perturbation adds a small gravitational acceleration, it's equivalent to adding a small mass term.But the perturbing acceleration a_moon = G M_m / D².This can be written as a_moon = G (M_m / D²).Comparing to the unperturbed acceleration a0 = G M_p / R².So, the ratio of the perturbing acceleration to the unperturbed acceleration is (M_m / D²) / (M_p / R²) = (M_m / M_p) * (R² / D²).Since the perturbation is small, this ratio is small.Now, to find the change in the orbital period, we can consider the perturbation as a small change in the effective gravitational parameter.Let me denote the unperturbed period as T0 = 2π sqrt(R³ / (G M_p)).The perturbed period T = T0 + ΔT.We can express the perturbed period as:T = 2π sqrt(R³ / (G (M_p + δM)))Where δM is a small correction to the mass due to the perturbation.But how does δM relate to the perturbing acceleration?Wait, the perturbing acceleration a_moon = G M_m / D².This can be thought of as an additional gravitational acceleration, so the effective gravitational acceleration becomes a0 + a_moon = G (M_p / R² + M_m / D²).But the orbital period depends on the gravitational parameter, which is G M_p for the planet. The perturbation adds a term G M_m / D², which can be considered as an additional gravitational parameter.Wait, perhaps we can write the effective gravitational parameter as G (M_p + M_m (R² / D²)).But that might not be directly accurate because the moon's gravity is not acting at the same point as the planet's gravity.Alternatively, perhaps we can consider the perturbation as a small change in the effective gravitational acceleration, which would lead to a small change in the orbital velocity and hence the period.Let me denote the unperturbed acceleration as a0 = G M_p / R².The perturbed acceleration is a = a0 + a_moon = G M_p / R² + G M_m / D².The orbital velocity is v = sqrt(a R).So, the perturbed velocity v = sqrt( (G M_p / R² + G M_m / D²) R ) = sqrt( G M_p / R + G M_m R / D² )But since the perturbation is small, we can expand this in a Taylor series.Let me write v = sqrt( G M_p / R + δa R ), where δa = G M_m / D².So, v ≈ sqrt( G M_p / R ) * sqrt(1 + (δa R) / (G M_p / R)) )Simplify the term inside the square root:(δa R) / (G M_p / R) = (G M_m / D² * R) / (G M_p / R) = (M_m / D²) * R² / M_pSo, v ≈ sqrt( G M_p / R ) * [1 + (1/2) (M_m R²) / (M_p D²) ]Therefore, the perturbed velocity v ≈ v0 [1 + (1/2) (M_m R²) / (M_p D²) ]Where v0 = sqrt( G M_p / R ) is the unperturbed velocity.Now, the orbital period is T = 2π R / v.So, substituting the perturbed velocity:T ≈ 2π R / [ v0 (1 + (1/2) (M_m R²) / (M_p D²) ) ]≈ 2π R / v0 [1 - (1/2) (M_m R²) / (M_p D²) ]Because 1 / (1 + x) ≈ 1 - x for small x.But 2π R / v0 is the unperturbed period T0.So, T ≈ T0 [1 - (1/2) (M_m R²) / (M_p D²) ]Therefore, the change in period ΔT = T - T0 ≈ - (1/2) T0 (M_m R²) / (M_p D² )So, ΔT ≈ - (1/2) T0 (M_m R²) / (M_p D² )But let's express R in terms of the planet's radius and altitude: R = R_p + h.So, substituting back:ΔT ≈ - (1/2) T0 (M_m (R_p + h)² ) / (M_p D² )But T0 itself is 2π sqrt( R³ / (G M_p) )So, substituting T0:ΔT ≈ - (1/2) * 2π sqrt( R³ / (G M_p) ) * (M_m (R_p + h)² ) / (M_p D² )Simplify:ΔT ≈ - π sqrt( R³ / (G M_p) ) * (M_m (R_p + h)² ) / (M_p D² )But R = R_p + h, so R³ = (R_p + h)³.Therefore:ΔT ≈ - π sqrt( (R_p + h)³ / (G M_p) ) * (M_m (R_p + h)² ) / (M_p D² )Simplify the expression:ΔT ≈ - π (R_p + h)² * sqrt( (R_p + h)³ / (G M_p) ) * M_m / (M_p D² )Combine the terms:= - π M_m / (M_p D² ) * (R_p + h)² * sqrt( (R_p + h)³ / (G M_p) )= - π M_m / (M_p D² ) * (R_p + h)² * (R_p + h)^(3/2) / sqrt(G M_p)= - π M_m / (M_p D² ) * (R_p + h)^(7/2) / sqrt(G M_p)= - π M_m / (M_p D² ) * (R_p + h)^(7/2) / (G^(1/2) M_p^(1/2))= - π M_m (R_p + h)^(7/2) / (G^(1/2) M_p^(3/2) D² )But this seems a bit complicated. Let me check my steps again.Wait, perhaps I made a mistake in the expansion. Let me go back.When I had v ≈ v0 [1 + (1/2) (M_m R²) / (M_p D²) ], then T = 2π R / v ≈ 2π R / (v0 (1 + x)) ≈ 2π R / v0 (1 - x), where x = (1/2) (M_m R²) / (M_p D²).But 2π R / v0 is T0, so T ≈ T0 (1 - x).Therefore, ΔT = T - T0 ≈ - T0 x = - T0 (1/2) (M_m R²) / (M_p D² )So, ΔT ≈ - (1/2) T0 (M_m R²) / (M_p D² )Expressing T0 as 2π sqrt( R³ / (G M_p) ), we have:ΔT ≈ - (1/2) * 2π sqrt( R³ / (G M_p) ) * (M_m R²) / (M_p D² )Simplify:ΔT ≈ - π sqrt( R³ / (G M_p) ) * (M_m R²) / (M_p D² )= - π M_m R² sqrt( R³ ) / (G^(1/2) M_p^(3/2) D² )= - π M_m R^(7/2) / (G^(1/2) M_p^(3/2) D² )But R = R_p + h, so substituting:ΔT ≈ - π M_m (R_p + h)^(7/2) / (G^(1/2) M_p^(3/2) D² )Hmm, that seems correct, but let me check the dimensions to ensure.The period T has dimensions of time, [T] = T.The expression for ΔT:M_m has dimensions of mass, [M].(R_p + h) has dimensions of length, [L].G has dimensions [L³ M⁻¹ T⁻²].M_p has dimensions [M].D has dimensions [L].So, numerator: M_m (R_p + h)^(7/2) = M L^(7/2)Denominator: sqrt(G) = (L³ M⁻¹ T⁻²)^(1/2) = L^(3/2) M^(-1/2) T^(-1)M_p^(3/2) = M^(3/2)D² = L²So, overall denominator: L^(3/2) M^(-1/2) T^(-1) * M^(3/2) * L² = L^(3/2 + 2) M^(-1/2 + 3/2) T^(-1) = L^(7/2) M^(1) T^(-1)So, numerator / denominator: (M L^(7/2)) / (L^(7/2) M T^(-1)) ) = TSo, the dimensions check out. Good.Therefore, the change in orbital period is:ΔT ≈ - (π M_m (R_p + h)^(7/2)) / ( sqrt(G) M_p^(3/2) D² )Alternatively, we can write sqrt(G) as G^(1/2), so:ΔT ≈ - π M_m (R_p + h)^(7/2) / (G^(1/2) M_p^(3/2) D² )But perhaps we can express this in terms of T0.Since T0 = 2π sqrt( (R_p + h)³ / (G M_p) ), we can write sqrt(G M_p) = 2π (R_p + h)^(3/2) / T0.Wait, let's solve for sqrt(G):sqrt(G) = (2π (R_p + h)^(3/2)) / (T0 sqrt(M_p))Wait, no, let's see:T0 = 2π sqrt( R³ / (G M_p) )So, T0² = 4π² R³ / (G M_p )Therefore, G M_p = 4π² R³ / T0²So, sqrt(G M_p) = 2π R^(3/2) / T0Therefore, sqrt(G) = (2π R^(3/2)) / (T0 sqrt(M_p))So, substituting back into ΔT:ΔT ≈ - π M_m R^(7/2) / ( (2π R^(3/2) / (T0 sqrt(M_p))) * M_p^(3/2) D² )Simplify denominator:(2π R^(3/2) / (T0 sqrt(M_p))) * M_p^(3/2) D² = 2π R^(3/2) M_p^(3/2) / (T0 sqrt(M_p)) ) * D²= 2π R^(3/2) M_p^( (3/2 - 1/2) ) / T0 * D²= 2π R^(3/2) M_p^(1) / T0 * D²So, ΔT ≈ - π M_m R^(7/2) / (2π R^(3/2) M_p / T0 D² )Simplify numerator and denominator:= - π M_m R^(7/2) * T0 D² / (2π R^(3/2) M_p )Cancel π:= - M_m R^(7/2 - 3/2) T0 D² / (2 M_p )= - M_m R² T0 D² / (2 M_p )Wait, that can't be right because the dimensions don't match. Wait, let me check.Wait, I think I made a mistake in the substitution.Let me try another approach. Let's express everything in terms of T0.We have T0 = 2π sqrt( R³ / (G M_p) )So, sqrt(G M_p) = 2π R^(3/2) / T0Therefore, G = (4π² R³) / (T0² M_p )So, sqrt(G) = (2π R^(3/2)) / (T0 sqrt(M_p))Now, substitute sqrt(G) into ΔT:ΔT ≈ - π M_m R^(7/2) / ( (2π R^(3/2) / (T0 sqrt(M_p))) * M_p^(3/2) D² )Simplify denominator:(2π R^(3/2) / (T0 sqrt(M_p))) * M_p^(3/2) D² = 2π R^(3/2) M_p^(3/2) / (T0 sqrt(M_p)) ) * D²= 2π R^(3/2) M_p^( (3/2 - 1/2) ) / T0 * D²= 2π R^(3/2) M_p^(1) / T0 * D²So, ΔT ≈ - π M_m R^(7/2) / (2π R^(3/2) M_p / T0 D² )Simplify numerator and denominator:= - π M_m R^(7/2) * T0 D² / (2π R^(3/2) M_p )Cancel π:= - M_m R^(7/2 - 3/2) T0 D² / (2 M_p )= - M_m R² T0 D² / (2 M_p )Wait, that can't be right because the units don't make sense. Let me check the exponents.Wait, R^(7/2) divided by R^(3/2) is R^(4/2) = R².So, yes, R².But then, ΔT ≈ - M_m R² T0 D² / (2 M_p )Wait, but D is in the denominator in the original expression, so in the substitution, D² is in the numerator, which seems odd.Wait, no, in the original expression, D is in the denominator, so when we substitute, D² remains in the denominator.Wait, let me double-check the substitution.Original ΔT expression:ΔT ≈ - π M_m R^(7/2) / (G^(1/2) M_p^(3/2) D² )We substituted G^(1/2) = (2π R^(3/2)) / (T0 sqrt(M_p))So, denominator becomes:(2π R^(3/2) / (T0 sqrt(M_p))) * M_p^(3/2) D²= 2π R^(3/2) M_p^(3/2) / (T0 sqrt(M_p)) ) * D²= 2π R^(3/2) M_p^( (3/2 - 1/2) ) / T0 * D²= 2π R^(3/2) M_p^(1) / T0 * D²So, denominator is 2π R^(3/2) M_p / T0 * D²Therefore, ΔT ≈ - π M_m R^(7/2) / (2π R^(3/2) M_p / T0 D² )= - π M_m R^(7/2) * T0 D² / (2π R^(3/2) M_p )Cancel π:= - M_m R^(7/2 - 3/2) T0 D² / (2 M_p )= - M_m R² T0 D² / (2 M_p )Wait, but D² is in the numerator here, which contradicts the original expression where D² was in the denominator. So, I must have made a mistake in the substitution.Wait, no, in the substitution, D² was in the denominator, so when we moved it to the numerator, it became D² in the numerator.But that leads to ΔT being proportional to D², which doesn't make physical sense because a larger D should lead to a smaller perturbation, hence a smaller ΔT.So, I must have made a mistake in the substitution.Wait, let's go back.Original expression:ΔT ≈ - π M_m (R_p + h)^(7/2) / (G^(1/2) M_p^(3/2) D² )We can express G^(1/2) as (G)^(1/2) = (G M_p)^(1/2) / M_p^(1/2)But from T0 = 2π sqrt( R³ / (G M_p) ), we have G M_p = 4π² R³ / T0²So, (G M_p)^(1/2) = 2π R^(3/2) / T0Therefore, G^(1/2) = (2π R^(3/2) / T0 ) / M_p^(1/2)So, substituting back:ΔT ≈ - π M_m R^(7/2) / ( (2π R^(3/2) / (T0 M_p^(1/2)) ) * M_p^(3/2) D² )Simplify denominator:(2π R^(3/2) / (T0 M_p^(1/2)) ) * M_p^(3/2) D² = 2π R^(3/2) M_p^( (3/2 - 1/2) ) / T0 * D²= 2π R^(3/2) M_p^(1) / T0 * D²So, ΔT ≈ - π M_m R^(7/2) / (2π R^(3/2) M_p / T0 D² )= - π M_m R^(7/2) * T0 D² / (2π R^(3/2) M_p )Cancel π:= - M_m R^(7/2 - 3/2) T0 D² / (2 M_p )= - M_m R² T0 D² / (2 M_p )Hmm, same result. But as I said, this suggests ΔT is proportional to D², which is counterintuitive because a larger D should lead to a smaller perturbation.Wait, perhaps I made a mistake in the initial assumption. Let me think again.When I considered the perturbing acceleration, I assumed it was in the same direction as the planet's gravity, but in reality, the moon's gravity can be in any direction, so the perturbation could be either increasing or decreasing the net gravitational force, leading to a change in the orbital period.But perhaps a better approach is to consider the perturbation as a small change in the effective gravitational parameter, which would lead to a change in the orbital period.Alternatively, maybe I should use the concept of the Hill sphere or consider the perturbation as a tidal force, but that might be more complex.Wait, another idea: The perturbation causes a small change in the effective gravitational acceleration, which in turn causes a small change in the orbital velocity, leading to a small change in the orbital period.Let me denote the unperturbed acceleration as a0 = G M_p / R².The perturbing acceleration is a_moon = G M_m / D².Assuming that the perturbing acceleration is small compared to a0, we can write the total acceleration as a = a0 + a_moon.The orbital velocity is v = sqrt(a R).So, the perturbed velocity v ≈ sqrt( (a0 + a_moon) R ) ≈ sqrt(a0 R) + (1/(2 sqrt(a0 R))) * a_moon R.= v0 + (a_moon R) / (2 v0 )Where v0 = sqrt(a0 R) is the unperturbed velocity.Therefore, the change in velocity Δv ≈ (a_moon R) / (2 v0 )Now, the orbital period is T = 2π R / v.So, the change in period ΔT can be found by differentiating T with respect to v:ΔT ≈ - (2π R) / v0² * Δv= - (2π R) / v0² * (a_moon R) / (2 v0 )= - (2π R) * a_moon R / (2 v0³ )= - π R² a_moon / v0³Now, a_moon = G M_m / D²v0³ = (G M_p / R )^(3/2 )So, substituting:ΔT ≈ - π R² (G M_m / D² ) / ( (G M_p / R )^(3/2 ) )Simplify:= - π R² G M_m / D² * R^(3/2) / (G^(3/2) M_p^(3/2) )= - π M_m R^(7/2) / (G^(1/2) M_p^(3/2) D² )Which matches the earlier result.So, despite the counterintuitive dependence on D², the math seems consistent.But physically, as D increases, the perturbation decreases, so ΔT should become smaller in magnitude. However, in our expression, ΔT is proportional to 1/D², which is correct because as D increases, the perturbation decreases, leading to a smaller |ΔT|.Wait, in the expression, ΔT is negative and proportional to -1/D², which makes sense because as D increases, the perturbation becomes smaller, so the change in period becomes less negative (i.e., the period decreases less or increases more, depending on the sign).Wait, actually, the sign depends on the direction of the perturbation. If the moon's gravity is adding to the planet's gravity, it would cause the space station to orbit faster, decreasing the period. If it's opposing, it would cause a slower orbit, increasing the period. But since we're taking the perturbation as a small acceleration, the sign would depend on the relative direction.But since the problem states that the perturbation is small and to treat it as a first-order correction, perhaps we can take the magnitude and note that the period changes by an amount proportional to M_m / D².Therefore, the change in orbital period is:ΔT ≈ - (π M_m (R_p + h)^(7/2)) / ( sqrt(G) M_p^(3/2) D² )Alternatively, expressing in terms of T0:Since T0 = 2π sqrt( (R_p + h)³ / (G M_p) ), we can write sqrt(G M_p) = 2π (R_p + h)^(3/2) / T0So, sqrt(G) = (2π (R_p + h)^(3/2)) / (T0 sqrt(M_p))Substituting back into ΔT:ΔT ≈ - π M_m (R_p + h)^(7/2) / ( (2π (R_p + h)^(3/2) / (T0 sqrt(M_p))) * M_p^(3/2) D² )Simplify denominator:= (2π (R_p + h)^(3/2) / (T0 sqrt(M_p))) * M_p^(3/2) D²= 2π (R_p + h)^(3/2) M_p^( (3/2 - 1/2) ) / T0 * D²= 2π (R_p + h)^(3/2) M_p / T0 * D²So, ΔT ≈ - π M_m (R_p + h)^(7/2) / (2π (R_p + h)^(3/2) M_p / T0 D² )= - M_m (R_p + h)^(7/2 - 3/2) T0 D² / (2 M_p )= - M_m (R_p + h)² T0 D² / (2 M_p )Wait, again, this suggests ΔT is proportional to D², which is problematic because as D increases, the perturbation should decrease, but this suggests the opposite.Wait, perhaps I made a mistake in the substitution. Let me check.Wait, in the denominator, we have D², so when we move it to the numerator, it becomes D² in the denominator.Wait, no, in the substitution, D² is in the denominator, so when we take the reciprocal, it becomes D² in the denominator.Wait, perhaps I should leave the expression as is, without substituting T0, because expressing it in terms of T0 might complicate the expression unnecessarily.So, the final expression for ΔT is:ΔT ≈ - (π M_m (R_p + h)^(7/2)) / ( sqrt(G) M_p^(3/2) D² )Alternatively, we can factor out T0:Since T0 = 2π sqrt( (R_p + h)³ / (G M_p) ), we can write:sqrt(G M_p) = 2π (R_p + h)^(3/2) / T0So, sqrt(G) = (2π (R_p + h)^(3/2)) / (T0 sqrt(M_p))Substituting into ΔT:ΔT ≈ - π M_m (R_p + h)^(7/2) / ( (2π (R_p + h)^(3/2) / (T0 sqrt(M_p))) * M_p^(3/2) D² )= - π M_m (R_p + h)^(7/2) * T0 sqrt(M_p) / (2π (R_p + h)^(3/2) M_p^(3/2) D² )Simplify:= - M_m (R_p + h)^(7/2 - 3/2) T0 / (2 M_p^( (3/2 - 1/2) ) D² )= - M_m (R_p + h)² T0 / (2 M_p D² )So, ΔT ≈ - (M_m (R_p + h)² T0 ) / (2 M_p D² )This seems more reasonable because now ΔT is proportional to 1/D², which makes sense.Therefore, the change in orbital period is:ΔT ≈ - (M_m (R_p + h)² T0 ) / (2 M_p D² )So, summarizing:1. The orbital period of the binary star system is T = 2π sqrt( d³ / (G (M1 + M2)) )2. The change in the orbital period of the space station due to the moon's perturbation is ΔT ≈ - (M_m (R_p + h)² T0 ) / (2 M_p D² )Where T0 is the unperturbed orbital period of the space station.I think this makes sense now. The negative sign indicates that if the moon's gravity is adding to the planet's gravity (i.e., pulling in the same direction), the period decreases, and if it's opposing, the period increases. But since the perturbation is small, the magnitude is given by the above expression.So, to recap:For the binary star system, the orbital period is derived using Newton's law of gravitation and centripetal force, leading to Kepler's third law.For the space station, the perturbation due to the moon is treated as a small correction, leading to a change in the orbital period proportional to the moon's mass, the square of the space station's orbital radius, and inversely proportional to the planet's mass and the square of the moon's distance.</think>

🧠:<think>To determine the number of students representing Norman and Anglo-Saxon soldiers, I first need to establish the historical ratio of soldiers. The Normans had approximately 7,000 soldiers, and the Anglo-Saxons had around 6,000 soldiers. This gives a ratio of 7:6.Next, I'll calculate the total number of parts in the ratio by adding 7 and 6, which equals 13 parts. Since there are 260 students participating, each part of the ratio corresponds to 260 divided by 13, which is 20 students per part.Using this, the number of Norman soldiers will be 7 parts multiplied by 20, resulting in 140 students. Similarly, the number of Anglo-Saxon soldiers will be 6 parts multiplied by 20, totaling 120 students.For the battlefield grid, I know that the total number of cells must equal the total number of soldiers, which is 260. The grid is rectangular with a length that is 5 cells longer than its width. Let's denote the width as ( w ) cells, making the length ( w + 5 ) cells. The area of the grid is then ( w times (w + 5) = 260 ).Expanding this equation gives ( w^2 + 5w - 260 = 0 ). Solving this quadratic equation using the quadratic formula, I find that the width is 13 cells. Consequently, the length is ( 13 + 5 = 18 ) cells.Therefore, the battlefield grid should be 13 cells wide and 18 cells long to accommodate all 260 soldiers.</think>

🧠:<think>Okay, so I have this problem about a circus trainer who needs to allocate medical care and training time to different animals. There are two parts to this problem. Let me try to tackle them one by one.Starting with the first part: The trainer has n animals, each requiring a specific amount of medical attention time, t_i, per week. The total available time is T hours. But there's a catch: each animal must receive at least 75% of their required time. The goal is to minimize the total deviation from the required times. Hmm, okay, so I need to set up a linear programming problem for this.First, let me define the variables. Let's say y_i is the actual time allocated to animal i. The required time is t_i, so the deviation for each animal would be |y_i - t_i|. But since we're dealing with linear programming, which can't handle absolute values directly, I remember that we can use a trick where we introduce two variables for each deviation: one for the positive deviation and one for the negative. So, let me define d_i^+ as the amount by which y_i exceeds t_i, and d_i^- as the amount by which y_i is less than t_i. Then, the total deviation would be the sum of d_i^+ and d_i^- for all i.But wait, the problem says we need to minimize the total deviation, so our objective function would be to minimize the sum of d_i^+ + d_i^- for all i. That makes sense.Now, the constraints. First, each animal must receive at least 75% of their required time. So, y_i >= 0.75 * t_i for each i. Second, the total time allocated can't exceed T, so the sum of y_i for all i must be less than or equal to T. Also, since y_i can't be negative, we have y_i >= 0.75 * t_i, which already covers the non-negativity.But wait, how do the d_i^+ and d_i^- fit into this? I think we need to relate y_i to t_i through these deviation variables. So, for each i, we can write y_i = t_i + d_i^+ - d_i^-. But since y_i must be at least 0.75 * t_i, we have t_i + d_i^+ - d_i^- >= 0.75 * t_i. Simplifying that, we get d_i^+ - d_i^- >= -0.25 * t_i. Hmm, that might complicate things. Alternatively, maybe it's better to express the deviation in terms of y_i.Wait, another approach: Since we're trying to minimize the total deviation, which is sum(|y_i - t_i|), and we can't have y_i less than 0.75 * t_i, perhaps we can express the deviation as (y_i - 0.75 t_i) if y_i >= t_i, or (t_i - y_i) if y_i < t_i. But since y_i can't be less than 0.75 t_i, the lower bound is 0.75 t_i, so the deviation is either y_i - t_i (if y_i > t_i) or t_i - y_i (if y_i < t_i). But since y_i can't go below 0.75 t_i, the maximum negative deviation is 0.25 t_i.But in linear programming, we can't have absolute values, so we need to split the deviation into two parts. So, for each i, we have:y_i = t_i + d_i^+ - d_i^-d_i^+, d_i^- >= 0But with the constraint that y_i >= 0.75 t_i, which translates to t_i + d_i^+ - d_i^- >= 0.75 t_i, so d_i^+ - d_i^- >= -0.25 t_i.But this is a bit messy. Maybe a better way is to express the deviation as (y_i - 0.75 t_i) if we consider that the minimum is 0.75 t_i. Wait, no, because the deviation is from t_i, not from 0.75 t_i.Alternatively, perhaps we can model the problem by considering that each y_i must be at least 0.75 t_i, but can be more. The deviation is how much we deviate from t_i, either above or below. But since we can't go below 0.75 t_i, the lower bound is fixed, so the deviation below is limited to 0.25 t_i.But maybe it's simpler to think of the deviation as (y_i - t_i) if y_i > t_i, and (t_i - y_i) if y_i < t_i, but with y_i >= 0.75 t_i. So, the deviation is either y_i - t_i or t_i - y_i, but y_i can't be less than 0.75 t_i.But in linear programming, we can't have conditional expressions, so we need to use the d_i^+ and d_i^- approach. So, for each i:y_i = t_i + d_i^+ - d_i^-d_i^+, d_i^- >= 0But with the constraint that y_i >= 0.75 t_i, which gives:t_i + d_i^+ - d_i^- >= 0.75 t_i=> d_i^+ - d_i^- >= -0.25 t_iBut since d_i^+ and d_i^- are both non-negative, this constraint might not be necessary because d_i^+ can be zero and d_i^- can be up to 0.25 t_i, but we need to ensure that y_i doesn't go below 0.75 t_i.Wait, maybe it's better to set y_i >= 0.75 t_i as a separate constraint, and then express the deviation as |y_i - t_i|, which we can't do directly, so we use d_i^+ and d_i^- as before.So, the constraints are:1. y_i >= 0.75 t_i for all i2. sum(y_i) <= T3. y_i = t_i + d_i^+ - d_i^- for all i4. d_i^+, d_i^- >= 0 for all iBut actually, we don't need to express y_i in terms of d_i^+ and d_i^-; instead, we can just use the deviations as variables. So, the deviation for each i is d_i = |y_i - t_i|, but since we can't have absolute values, we introduce d_i^+ and d_i^- such that y_i - t_i = d_i^+ - d_i^-, and d_i^+, d_i^- >= 0. Then, the total deviation is sum(d_i^+ + d_i^-).So, the objective function is to minimize sum(d_i^+ + d_i^-).Subject to:1. y_i >= 0.75 t_i for all i2. sum(y_i) <= T3. y_i - t_i = d_i^+ - d_i^- for all i4. d_i^+, d_i^- >= 0 for all iBut actually, since y_i - t_i = d_i^+ - d_i^-, we can write y_i = t_i + d_i^+ - d_i^-. So, substituting into the first constraint:t_i + d_i^+ - d_i^- >= 0.75 t_i=> d_i^+ - d_i^- >= -0.25 t_iBut since d_i^+ and d_i^- are non-negative, this inequality is automatically satisfied because d_i^+ >= 0 and -d_i^- >= -0.25 t_i, which implies d_i^- <= 0.25 t_i. So, we can include this as a constraint: d_i^- <= 0.25 t_i for all i.But wait, is that necessary? Because if d_i^- is allowed to be up to 0.25 t_i, then y_i can be as low as 0.75 t_i. So, maybe we can include that as a constraint to ensure that we don't have d_i^- exceeding 0.25 t_i.Alternatively, since y_i >= 0.75 t_i, and y_i = t_i + d_i^+ - d_i^-, we can rearrange to get d_i^+ - d_i^- >= -0.25 t_i. But since d_i^+ and d_i^- are non-negative, this is equivalent to d_i^- <= d_i^+ + 0.25 t_i. Hmm, not sure if that's helpful.Maybe it's better to just include y_i >= 0.75 t_i as a separate constraint and not worry about the d_i^- because the solver will handle it.So, putting it all together, the linear program is:Minimize sum(d_i^+ + d_i^-) for all iSubject to:1. y_i >= 0.75 t_i for all i2. sum(y_i) <= T3. y_i = t_i + d_i^+ - d_i^- for all i4. d_i^+, d_i^- >= 0 for all iBut actually, since y_i = t_i + d_i^+ - d_i^-, we can substitute this into the first constraint:t_i + d_i^+ - d_i^- >= 0.75 t_i=> d_i^+ - d_i^- >= -0.25 t_iBut since d_i^+ and d_i^- are non-negative, this is equivalent to d_i^- <= d_i^+ + 0.25 t_i. However, since d_i^- can't be negative, and d_i^+ can be zero, this constraint might not be necessary because the other constraints will handle it.Wait, maybe not. Let me think. If d_i^- is allowed to be up to 0.25 t_i, then y_i can be as low as 0.75 t_i. So, perhaps we should include d_i^- <= 0.25 t_i as a constraint to prevent y_i from going below 0.75 t_i.Alternatively, since y_i >= 0.75 t_i is already a constraint, and y_i = t_i + d_i^+ - d_i^-, then:t_i + d_i^+ - d_i^- >= 0.75 t_i=> d_i^+ - d_i^- >= -0.25 t_iWhich can be rewritten as d_i^- <= d_i^+ + 0.25 t_iBut since d_i^+ >= 0, this allows d_i^- to be up to 0.25 t_i if d_i^+ is zero. So, perhaps we can include d_i^- <= 0.25 t_i as a constraint to ensure that y_i doesn't go below 0.75 t_i.Alternatively, maybe it's redundant because the constraint y_i >= 0.75 t_i already enforces that.I think it's safer to include both y_i >= 0.75 t_i and y_i = t_i + d_i^+ - d_i^-, and let the solver handle the rest. So, the constraints are:1. y_i >= 0.75 t_i for all i2. sum(y_i) <= T3. y_i = t_i + d_i^+ - d_i^- for all i4. d_i^+, d_i^- >= 0 for all iAnd the objective is to minimize sum(d_i^+ + d_i^-).This should work. Now, solving this linear program would give the optimal allocation of y_i that minimizes the total deviation while satisfying the constraints.But wait, let me think about the dual problem or if there's a simpler way. Maybe using Lagrange multipliers, but since it's linear programming, the simplex method would be appropriate.Alternatively, maybe there's a way to express this without the d_i^+ and d_i^- variables. Let me consider that.If we don't use the deviation variables, the problem becomes minimizing sum(|y_i - t_i|) subject to y_i >= 0.75 t_i and sum(y_i) <= T.But since linear programming can't handle absolute values, we need to linearize it, which is why we introduced d_i^+ and d_i^-.So, I think the formulation I have is correct.Now, moving on to the second part: The trainer also needs to consider the joint health effects of medical care and training. Each animal has a health improvement function H_i(t_i, x_i) = a_i t_i^{0.5} + b_i x_i^{0.3}, where x_i is the training time, and a_i, b_i are positive constants. Given a total training time X, we need to maximize the sum of H_i for all i, subject to sum(x_i) = X.This is a constrained optimization problem. Since the functions are separable in x_i, we can use the method of Lagrange multipliers.Let me set up the Lagrangian:L = sum_{i=1}^n [a_i t_i^{0.5} + b_i x_i^{0.3}] + λ (X - sum_{i=1}^n x_i)Wait, but t_i is the medical care time, which was determined in the first part. So, t_i is fixed for each animal, and x_i is the variable we're optimizing here. So, the Lagrangian should be:L = sum_{i=1}^n [a_i t_i^{0.5} + b_i x_i^{0.3}] + λ (X - sum_{i=1}^n x_i)But since a_i t_i^{0.5} is a constant with respect to x_i, the derivative of L with respect to x_i will be the derivative of b_i x_i^{0.3} minus λ.So, taking the partial derivative of L with respect to x_i:dL/dx_i = 0.3 b_i x_i^{-0.7} - λ = 0Solving for x_i:0.3 b_i x_i^{-0.7} = λ=> x_i^{-0.7} = λ / (0.3 b_i)=> x_i = (0.3 b_i / λ)^{1/0.7}But 1/0.7 is approximately 1.4286, which is 10/7.So, x_i = (0.3 b_i / λ)^{10/7}But we also have the constraint sum(x_i) = X.So, substituting x_i into the constraint:sum_{i=1}^n (0.3 b_i / λ)^{10/7} = XLet me denote C = (0.3 / λ)^{10/7}, then:C sum_{i=1}^n b_i^{10/7} = X=> C = X / sum_{i=1}^n b_i^{10/7}Therefore, x_i = C b_i^{10/7} = (X / sum_{i=1}^n b_i^{10/7}) * b_i^{10/7}So, the optimal x_i is proportional to b_i^{10/7}.Wait, let me check the exponents again. We had x_i = (0.3 b_i / λ)^{10/7}, and then C = (0.3 / λ)^{10/7}, so x_i = C b_i^{10/7}.Yes, that seems correct.So, the optimal allocation is x_i proportional to b_i^{10/7}.But let me verify the derivative step again. The derivative of b_i x_i^{0.3} with respect to x_i is 0.3 b_i x_i^{-0.7}, which is correct. Setting this equal to λ gives x_i^{-0.7} = λ / (0.3 b_i), so x_i = (0.3 b_i / λ)^{1/0.7} = (0.3 b_i / λ)^{10/7}.Yes, that's correct.So, the optimal x_i is x_i = (0.3 b_i / λ)^{10/7}, and sum(x_i) = X, so we solve for λ in terms of X and the b_i's.Alternatively, we can express x_i as:x_i = k b_i^{10/7}, where k is a constant such that sum(k b_i^{10/7}) = X => k = X / sum(b_i^{10/7}).Therefore, the optimal allocation is x_i = (X / sum(b_j^{10/7})) * b_i^{10/7}.So, each x_i is proportional to b_i^{10/7}.Wait, but let me think about the exponents again. The exponent in the derivative was -0.7, so when we take the reciprocal, it's 1/0.7 ≈ 1.4286, which is 10/7. So, yes, the exponent is 10/7.Therefore, the optimal x_i is proportional to b_i^{10/7}.So, summarizing, for the second part, the optimal allocation is x_i = (X / sum(b_j^{10/7})) * b_i^{10/7}.I think that's correct.But let me double-check. If we have H_i = a_i t_i^{0.5} + b_i x_i^{0.3}, and we're maximizing sum(H_i) with sum(x_i) = X, then the marginal gain from x_i is 0.3 b_i x_i^{-0.7}, and we set these equal across all i, which gives x_i proportional to b_i^{10/7}.Yes, that makes sense because the marginal utility decreases as x_i increases, so we allocate more to the ones with higher b_i.So, in conclusion, for the first part, we set up a linear program with variables y_i, d_i^+, d_i^-, and minimize the total deviation subject to the constraints. For the second part, we use Lagrange multipliers and find that x_i is proportional to b_i^{10/7}.I think that's it. I might have missed some steps, but overall, that's the approach.</think>

🧠:<think>Okay, so I have this problem about the Glasgow Warriors' attendance at home games. It says that the attendance follows a quadratic sequence, and they've given me the first three attendances: 12,000; 13,500; and 15,600. I need to find the quadratic formula for the attendance sequence. Then, part two is about finding the cumulative attendance over the first n games, and specifically for 10 games. Hmm, okay, let me start with part one.First, I remember that a quadratic sequence has the form ( a_n = an^2 + bn + c ), where a, b, c are constants. So, for each game number n, the attendance is given by this formula. Since they've given me the first three terms, I can set up equations and solve for a, b, c.Let me write down the equations based on the given attendances:For n = 1: ( a(1)^2 + b(1) + c = 12,000 )Which simplifies to: ( a + b + c = 12,000 ) --- Equation 1For n = 2: ( a(2)^2 + b(2) + c = 13,500 )Which is: ( 4a + 2b + c = 13,500 ) --- Equation 2For n = 3: ( a(3)^2 + b(3) + c = 15,600 )Which is: ( 9a + 3b + c = 15,600 ) --- Equation 3Now, I have three equations:1. ( a + b + c = 12,000 )2. ( 4a + 2b + c = 13,500 )3. ( 9a + 3b + c = 15,600 )I need to solve this system of equations to find a, b, c.Let me subtract Equation 1 from Equation 2 to eliminate c:Equation 2 - Equation 1: (4a + 2b + c) - (a + b + c) = 13,500 - 12,000Simplify: 3a + b = 1,500 --- Equation 4Similarly, subtract Equation 2 from Equation 3:Equation 3 - Equation 2: (9a + 3b + c) - (4a + 2b + c) = 15,600 - 13,500Simplify: 5a + b = 2,100 --- Equation 5Now, I have two equations:4. ( 3a + b = 1,500 )5. ( 5a + b = 2,100 )Subtract Equation 4 from Equation 5 to eliminate b:(5a + b) - (3a + b) = 2,100 - 1,500Simplify: 2a = 600So, a = 300Now, plug a = 300 into Equation 4:3*(300) + b = 1,500900 + b = 1,500So, b = 1,500 - 900 = 600Now, plug a = 300 and b = 600 into Equation 1:300 + 600 + c = 12,000900 + c = 12,000So, c = 12,000 - 900 = 11,100So, the quadratic formula is:( a_n = 300n^2 + 600n + 11,100 )Wait, let me check if this works for the given terms.For n=1: 300*(1) + 600*(1) + 11,100 = 300 + 600 + 11,100 = 12,000 ✔️For n=2: 300*(4) + 600*(2) + 11,100 = 1,200 + 1,200 + 11,100 = 13,500 ✔️For n=3: 300*(9) + 600*(3) + 11,100 = 2,700 + 1,800 + 11,100 = 15,600 ✔️Perfect, that's correct.So, part one is done. The quadratic formula is ( a_n = 300n^2 + 600n + 11,100 ).Now, moving on to part two. The fan wants the cumulative attendance over the first n home games. So, cumulative attendance ( S_n ) is the sum of the first n terms of the quadratic sequence.I remember that the sum of a quadratic sequence can be found using the formula for the sum of a quadratic series. Alternatively, since each term is ( a_n = an^2 + bn + c ), the sum ( S_n ) would be the sum from k=1 to n of ( ak^2 + bk + c ). So, that can be split into three separate sums:( S_n = a sum_{k=1}^{n} k^2 + b sum_{k=1}^{n} k + c sum_{k=1}^{n} 1 )I know the formulas for these sums:1. ( sum_{k=1}^{n} k^2 = frac{n(n+1)(2n+1)}{6} )2. ( sum_{k=1}^{n} k = frac{n(n+1)}{2} )3. ( sum_{k=1}^{n} 1 = n )So, substituting these into the expression for ( S_n ):( S_n = a left( frac{n(n+1)(2n+1)}{6} right) + b left( frac{n(n+1)}{2} right) + c(n) )So, that's the general formula for the cumulative attendance in terms of a, b, c, and n.But since we already found a, b, c for the specific case, we can plug those values in to get a specific formula for ( S_n ).Given that a = 300, b = 600, c = 11,100.So, substituting:( S_n = 300 left( frac{n(n+1)(2n+1)}{6} right) + 600 left( frac{n(n+1)}{2} right) + 11,100(n) )Let me simplify each term step by step.First term: 300 * [n(n+1)(2n+1)/6]Simplify 300/6 = 50, so it becomes 50n(n+1)(2n+1)Second term: 600 * [n(n+1)/2]Simplify 600/2 = 300, so it becomes 300n(n+1)Third term: 11,100nSo, putting it all together:( S_n = 50n(n+1)(2n+1) + 300n(n+1) + 11,100n )Now, let's see if we can factor or simplify this further.Looking at the first two terms, both have n(n+1) as a factor. Let me factor that out:( S_n = n(n+1)[50(2n+1) + 300] + 11,100n )Compute inside the brackets:50(2n + 1) + 300 = 100n + 50 + 300 = 100n + 350So, now:( S_n = n(n+1)(100n + 350) + 11,100n )Let me expand n(n+1)(100n + 350):First, multiply n and (n+1):n(n+1) = n^2 + nThen, multiply by (100n + 350):(n^2 + n)(100n + 350) = n^2*(100n + 350) + n*(100n + 350)= 100n^3 + 350n^2 + 100n^2 + 350n= 100n^3 + (350n^2 + 100n^2) + 350n= 100n^3 + 450n^2 + 350nSo, now, the cumulative sum is:( S_n = 100n^3 + 450n^2 + 350n + 11,100n )Combine like terms:350n + 11,100n = 11,450nSo, ( S_n = 100n^3 + 450n^2 + 11,450n )Alternatively, we can factor out a common factor if possible. Let me see:All coefficients are divisible by 50? 100/50=2, 450/50=9, 11,450/50=229.So, factoring out 50:( S_n = 50(2n^3 + 9n^2 + 229n) )But I don't know if that's necessary. Alternatively, we can leave it as is.So, the formula for cumulative attendance is ( S_n = 100n^3 + 450n^2 + 11,450n ).Now, the question also asks for the cumulative attendance if the season has 10 home games. So, we need to compute ( S_{10} ).Let me compute each term step by step.First, compute each part:100n^3: 100*(10)^3 = 100*1000 = 100,000450n^2: 450*(10)^2 = 450*100 = 45,00011,450n: 11,450*10 = 114,500Now, add them together:100,000 + 45,000 = 145,000145,000 + 114,500 = 259,500So, the cumulative attendance over 10 home games is 259,500.Wait, let me cross-verify this with another method to ensure I didn't make a calculation error.Alternatively, I can compute each term individually:First, compute each a_n for n=1 to 10 and sum them up.But that would be tedious, but maybe I can compute a few terms to check.Wait, let me see:Alternatively, I can compute ( S_n ) using the formula:( S_n = 50n(n+1)(2n+1) + 300n(n+1) + 11,100n )Plugging n=10:First term: 50*10*11*21Compute 10*11=110, 110*21=2,310, 50*2,310=115,500Second term: 300*10*11=300*110=33,000Third term: 11,100*10=111,000Now, sum them: 115,500 + 33,000 = 148,500; 148,500 + 111,000 = 259,500Same result. So, that's correct.Alternatively, using the expanded formula:( S_n = 100n^3 + 450n^2 + 11,450n )For n=10:100*(1000) + 450*(100) + 11,450*(10) = 100,000 + 45,000 + 114,500 = 259,500Consistent again. So, that's correct.Therefore, the cumulative attendance over 10 home games is 259,500.So, summarizing:1. The quadratic formula for attendance is ( a_n = 300n^2 + 600n + 11,100 ).2. The cumulative attendance formula is ( S_n = 100n^3 + 450n^2 + 11,450n ), and for n=10, it's 259,500.Final AnswerThe quadratic formula for attendance is boxed{300n^2 + 600n + 11100}, and the cumulative attendance over 10 home games is boxed{259500}.</think>