Asked by Eden
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?
Answers
Answered by
Damon
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
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
Answered by
Eden
Thanks for the help, but it said it wasn't right.... I did 4.47*(6.38*10^3)
Is that right?
Is that right?
Answered by
Damon
earth radius = 6.38*10^6
not 10^3
not 10^3
Answered by
Damon
28.5 *10^6 = 2.85 * 10^7 meters
Answered by
Anonymous
this just isnt right...
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