For a spherical celestial object of radius R, the acceleration due to gravity g

Question

For a spherical celestial object of radius R, the acceleration due to gravity g at a distance x from the center of the object is g = g_0R^2/x^2, where g_0 is the acceleration due to gravity at the object's surface and x > R. For the moon, take g_0 = 1.63 m/s^2 and R = 3200 km. If a rock is released from rest at a height of 5R above the lunar surface, with what speed does the rock impact the moon? Its acceleration is a function of position and increases as the object falls. So do not use constant acceleration free-fall equations, but go back to basics.

Explanation / Answer

given acceleration

g = go * R^2 / x^2

g = 1.63*3200*3200*10^6 /x^2

g = 1.67*10^13/x^2

let K = 1.67x10^13

g = K/x^2

now acceleration g = vdv/dx

vdv/dx = K/x^2

vdv = K(1/x^2)dx

integrating this

limits for speed are from 0 to v and for x limits are from 5R to R we get

v^2 / 2 = K(-1/x)from(5R to R)

V^2 / 2 = K(1/R - 1/5R)

V = sqrt(2*K(4R/5R^2))

V = 2888.8 m/s

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