WHEN WORLDS COLLIDE f7x103 and masses of 5x10 are separated by 4x10? miles. They

Question

WHEN WORLDS COLLIDE f7x103 and masses of 5x10 are separated by 4x10? miles. They are initially at rest. How fast are they moving just before their surfaces colide. Assume that you can ignore any effects having to do with the existence of atmospheres and that noching significant exists in the space between the planets Answer: 6) Consider the same setup as before, only one planet is twice the mass of the other (their radi are still the same size). What is the speed of the more massive planet just before their surfaces collide? Answer:

Explanation / Answer

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*x^2 + 0.5*M2*y^2

or (-6.67*10^-11/(1.609*4*10^7) + 6.67*10^-11/(1.609*7*10^3*2))*5*10^24 = 0.5*x^2 + 0.5*y^2

using the two equations,

we get x=y=1.21*10^5 m/s

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*x^2 + 0.5*M2*y^2

or (-6.67*10^-11/(1.609*4*10^7) + 6.67*10^-11/(1.609*7*10^3*2))*2*5*10^24 = 0.5*2*x^2 + 0.5*y^2

Solving the two equations,. we get

x = 99330.9 m/s

y = 1.986*10^5 m/s

Hence speed of more massive planet = 99330.9 m/s

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