Posted by **Eden** on Sunday, October 3, 2010 at 7:48pm.

What is the distance from the Earth's center to a point outside the Earth where the gravitational acceleration due to the Earth is 1/20 of its value at the Earth's surface?

- Physics -
**Damon**, Sunday, October 3, 2010 at 7:56pm
m g = G m M / r^2

so

g = G M / r^2

g' = G M /R^2

g'/g = 1/r'^2 / 1/r^2

g'/g = r^2/r'^2

1/20 = r^2/r'^2

so r'/r = sqrt 20 = 4.47

so r' = 4.47 * earth radius

- Physics -
**Eden**, Sunday, October 3, 2010 at 8:04pm
Thanks for the help, but it said it wasn't right.... I did 4.47*(6.38*10^3)

Is that right?

- Physics -
**Damon**, Sunday, October 3, 2010 at 8:25pm
earth radius = 6.38*10^6

not 10^3

- Physics -
**Damon**, Sunday, October 3, 2010 at 8:26pm
28.5 *10^6 = 2.85 * 10^7 meters

- Physics -
**Anonymous**, Wednesday, November 17, 2010 at 3:25pm
this just isnt right...

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