# Object falling; acceleration based on x

## Homework Statement

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 = (g0)(R^2)/(x^2), where g0 is the acceleration due to gravity at the object's surface and x > R. For the moon, take g0 = 1.63 m/s2 and R = 3200 km. If a rock is released from rest at a height of 3R above the lunar surface, with what speed does the rock impact the moon? Hint: 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.

## Homework Equations

.5(v^2-v0^2) = (g0)(R^2)(1/x-1/x0)

## The Attempt at a Solution

i said that v0=0 and solved for v, x=3R x0=R, i got .899km/s. apparently that is incorrect.

gneill
Mentor
What's the difference between height and radial distance?

2R, so would i plug 2R in as x and 0 for x0?

gneill
Mentor
2R, so would i plug 2R in as x and 0 for x0?

What's the radial distance to the surface of the Moon?

What's the radial distance to an object at height 3R above the Moon's surface?

What's the radial distance to the surface of the Moon?

R

What's the radial distance to an object at height 3R above the Moon's surface?

4R

I see now, i think.
so x0=4R, x=R right?

i got 2.797km/s is that the correct answer?

gneill
Mentor
It looks good.

gneill
Mentor
Does the system care if you enter the result as km/sec or m/sec?

it wants the answer in km/s

gneill
Mentor
Is it fussy about significant figures?

i got it, i think i plugged it into the calculator wrong.

thanks for all the help