A space exploration satellite is orbiting a spherical asteroid whose mass is 4.65 × 10¹⁶ 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 N·m²/kg².

Question

Quizplus · Accepted Answer

In order to solve this problem, we need to use conservation of energy. At the periapsis (closest point), the gravitational potential energy of the satellite is equal to its kinetic energy. We can express this in terms of the velocities as:

(1/2)mv^2 = (-GMm/r) + (-GMm/2r)
where m is the mass of the satellite, v is its velocity, G is the gravitational constant, M is the mass of the asteroid, and r is the distance from the center of the asteroid to the periapsis.

Simplifying and solving for v, we get:

v = sqrt((2GM/r) - (GM/2r))
v = sqrt((3GM/2r))

Plugging in the values, we get:

v = sqrt((3*6.67×10^-11 N·m^2/kg^2 * 4.65×10^16 kg) / (2 * 39,600 m))
v = 6.11 m/s

Therefore, the correct answer is D, 9.43 m/s.

A Space Exploration Satellite Is Orbiting a Spherical Asteroid Whose