Question

Two heavenly bodies both with radii of 4×10³ miles and masses of 8×10²⁴ kg are separated by 4×10⁷ miles. They are initially at rest. Assume that you can ignore any effects having to do with the existence of atmospheres and that nothing significant exists in the space between the planets.

Consider the same setup as before, only one planet is twice the mass of the other (their radii are still the same size). What is the speed of the more massive planet just before their surfaces collide?

Answer 1

Part a)

let the speed of one body be 'x'

and

the speed of other body be 'y'.

Applying law of conservation of momentum,

M₁*x = M₂*y

since,

M₁ = M₂, & x = y

Also,

applying law of conservation of energy,

loss in potential energy = gain in kinetic energy

or

-G*M₁*M₂/D + G*M₁*M₂/(R₁+R₂) = 0.5*M₁*x² + 0.5*M₂*y²

[(-6.67*10^-11)/(1.609*4*10⁷) + (6.67*10^-11)/(1.609*4*10³*2)]*8*10²⁴ = 0.5*x² + 0.5*y²

using the two equations,

we get x = y = 2.037*10⁵ m/s

Part b)

Apply law of conservation of momentum,

M₁*x = M₂*y

let M₁ = 2*M₂, so we get,

2x = y

conserving energy,

-G*M₁*M₂/D + G*M₁*M₂/(R₁+R₂) = 0.5*M₁*x² + 0.5*M₂*y²

or (-6.67*10^-11/(1.609*4*10^7) + 6.67*10^-11/(1.609*4*10^3*2))*2*8*10^24 = 0.5*2*x^2 + 0.5*(2x)^2

Solving the two equations,. we get

x = 1.66*10^5 m/s

y = 3.33*10^5 m/s

Hence speed of more massive planet = 1.66*10⁵ m/s