Question

An asteroid of mass m is in a perfectly circular orbit of radius d about a planet of mass M. You may assume that M>>m so that M is completely stationary.

a. Determine an expression for the orbital period of the asteroid.

(I already completed this step. Answer is T=2pi(r^3/2)/sqrt(GM)

b. The asteroid is suddenly stopped in its tracks and allowed to fall freely towards the planet. Determine the time it takes to collide with the planet (assumed to have a radius of zero) and express that time in terms of your answer to part a. (Hint: you will need to determine dr/dt (r) and ultimately trigonometric substitutions will be useful.)

I am stuck on part b. The answer is supposed to be t=T/(4sqrt(2)) but I can't get there. I don't have any trig stuff to substitute with.

Answer 1

answer b)

let r= radius of the asteroid around the planet

time taken to reach the planet=t

and time taken by the asteroid to revolve around the planet of radius r =T

let us consider 2t'=T', which is the time period of collision between the planet and the asteroid

now T/T'=(r/r')^3/2

r'=1/2r

T'=T(1/2)^3/2

we have T'=2t'

so 2t'=T(1/2)^3/2

t'=T(1/2)^3/2*1/2

t'=T$\sqrt{}$2/8

t'=T$\sqrt{}$2/4*$\sqrt{}$2*$\sqrt{}$2

t'=T/4$\sqrt{}$2

so the answer is t'=T/4$\sqrt{}$2