Object falling; acceleration based on x

[Sheldinoh]

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]

What's the difference between height and radial distance?

 

[Sheldinoh]

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

 

[gneill]

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?

 

[Sheldinoh]

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?

 

[Sheldinoh]

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

 

[gneill]

It looks good.

 

[Sheldinoh]

ok i just tried it and it said i have the incorrect answer, and it gave me this link http://www-physics.ucsd.edu/students/courses/winter2011/physics2a/Tipler6-2.P.122.pdf

do you have any ideas on where i might have messed up?

 

[gneill]

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

 

[Sheldinoh]

it wants the answer in km/s

 

[gneill]

Is it fussy about significant figures?

 

[Sheldinoh]

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

 

[Sheldinoh]

thanks for all the help