A space exploration satellite is orbiting a spherical asteroid whose mass is 4.6

Question

A space exploration satellite is orbiting a spherical asteroid whose mass is 4.65 × 10^16 kg and whose radius is 39,600 m, at an altitude of 12,400 m above the surface of the asteroid. In order to make a soft landing, Mission Control sends it a signal to fire a short burst of its retro rockets to change its speed to one that will put the satellite in an elliptical orbit with a periapsis (the distance of closest approach, as measured from the center of the asteroid) equal to the radius of the asteroid. What is the speed of the satellite when it reaches the surface of the asteroid? G= 6.67 x 10^-11 nm^2/kg^2

answer is 9.43 m/s

can you show how to get the answer

Explanation / Answer

:  We will use K+U [energy conservation] to solve this. In orbit K = 1/2*m*v1^2 and U = -GMm/r
where r = 39600+12400 m = 52000m v1 can be determined from GMm/r^2 = m*v1^2/r or v1^2 = GM/r

Now at the surface U = -GMm/R where R = 39600m and K = 1/2*m*v^2. Our goal is to find v..
So

setting K+U orbit = K+U surface we get 1/2*m*GM/r - GMm/r = 1/2*m*v^2 - GMm/R. Now simplifying (mass m is

not needed) we get v^2 - GM/R = GM/r - 2*GM/r

So v = sqrt( GM/R +GM/r -2*GM/r) = sqrt( GM/R -GM/r) = sqrt (6.67x10^-11*4.65x10^16*(1/39600 - 1/52000)

= 4.32 m/s

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