Question

Two planets are separated in space by some distance d, each orbiting around their center of mass in the middle of them. They both have the same mass m = 4.81 × 10^20 kg and are rotating with a constant ω = 1.25 × 10^−10 rad/s. How far apart are they (d)?

The answer is 1.6 x 10^10 meters.

Answer 1

The two planets are orbiting around each other about their centre of mass. Let the distances between the planets from the centre of mass of their system be d₁ and d_2.

The distance between the two planets would now be equal to d = d₁ + d₂

Let the centre of mass be at x = 0

therefore the centre of mass equation will be:

$\vec{X} = \frac{M_1d_1 + M_2(-d_2)}{M}$

$0 = \frac{M_1d_1 - M_2d_2}{M}$

therefore, M₁d₁ = M₂d₂

and d = d₁ + d₂

and so d = d₁ + (d₁M₁/M₂)

writing it in terms of d₂ we get,

$d_2 = \frac{M_1}{M_1 +M_2}d$.......................(1)

and we also know that for a system like this, the Gravitational Force provided by M₁ gives the necessary centripetal force to M₂ to remain in the orbit.

$F_{grav} = GM_1M_2/d^2$............................(2)

$F_{centripetal} = M_2\omega^2 d_2$..........................(3)

Notice the distance in the centripetal force is not d since M₂ revolves around the centre of mass, not M_1.

therefore,

so equating equating (1) and (2), we get:

$GM_1M_2/d^2 = M_2\omega^2 d_2$

or $\frac{GM_1M_2}{d^2M_2d_2} = \omega^2$

substitute the value of d₂ from equation (1)

$\frac{GM_1M_2}{d^2M_2 \frac{M_1d}{M_1 +M_2}} = \omega^2$

$\frac{G(M_1 +M_2)}{d^3} = \omega^2$

or $d^3 = \frac{G(M_1 +M_2)} {\omega^2}$

but M₁ = M₂

therefore, $d^3 = \frac{G2M} {\omega^2}$

plug in the value of G = 6.67 x 10^-11 m³kg^-1s^-2 , M = 4.81 x 10²⁰kg and angular velocity = 1.25 x 10^-10 rad/s to find the value of d.

d = 1.601 x 10¹⁰ m