AP Physics Question about gravitational field

Question

Two planets P₁ and P₂ orbit around a star S in circular orbits with speeds v₁ = 45.8 km/s, and v₂ = 56.8 km/s respectively.

(a) If the period of the first planet P₁ is 740 years what is the mass, in kg, of the star it orbits around? (unit: kg)

(b) Determine the orbital period, in years, of P₂. (units: years)

Thank you.

Answer 1

Note:

Below we will use 10⁷π as an approximation to the number of seconds in an Earth year.

Let

v₁ = 45.8 km/s = 45800 m/s be the speed of P₁

v₂ = 56.8 km/s = 56800 m/s be the speed of P₂

r₁ (unknown) be the radius of P₁

r₂ (unknown) be the radius of P₂

T₁ = 740 years = 740 × 10⁷π s be orbital period of P₁

T₂ (unknown) be the orbital period of P₂

M (unknown) be the mass of the star

G = 6.674×10^-11 m³⋅kg^-1s^-2 be the gravitational constant

First consider an arbitrary planet P with

speed v, radius r, period T.

Since the planet goes around a circle of radius r in time T,

2πr = vT (1)

The planet P experiences gravitation acceleration from Newton's Gravitational Law:

GM/r²

That causes centripetal acceleration

v²/r

Therefore the two are equal

v²/r = GM/r²

Simplifying:

v²r = GM (2)

(a)

Apply (1) to planet P₁ to express r₁:

r₁ = v₁T₁/2π (3)

Apply (2) to planet P₁ to express M:

M = v₁²r₁/G (4)

Substitute (3) into (4) and simplify

M = v₁³T₁/2πG (5)

Using actual numbers

M = 45800³ × 740 × 10⁷π / 2π × 6.674×10^-11 = 5.3×10³³kg

That is about 2700 solar masses.

(b)

Apply (2) to planet P₂ to express r₂:

r₂ = GM/v₂² (6)

Apply (1) to planet P₂ to express T₂:

T₂ = 2πr₂/v₂ (7)

Substitute (6) into (7)

T₂ = 2πGM/v₂³ (8)

Substitute (5) into (8) and simplify

T₂ = T₁(v₁/v₂)³

Using actual numbers

T₂ = 740 × (45.8/56.8)³ = 388 years

Answer 2

Great question Kat!

Hint: The key is to set the centripetal and gravitational forces equal to each other.

Solution: https://imgur.com/a/D3GdTax