Posted by **johnny** on Saturday, December 6, 2008 at 6:43pm.

At what altitude above Earth's surface would the gravitational acceleration be 3.0 m/s2?

- physics -
**Damon**, Saturday, December 6, 2008 at 7:40pm
I am not going to do all this arithmetic

F/m = g apparent = 3 = G Mearth/(Rearth +h)^2

where we know

9.8 = G Mearth/Rearth^2

so

9.8/3 = (Rearth+h)^2 / Rearth^2

- physics -
**GK**, Saturday, December 6, 2008 at 7:51pm
Let:

m = a small mass

M = mass of the earth

r = distance from the center of mass m to the earth's center.

Constants:

G = 6.67x10^-11 m^3/kg.s^2.

M = 5.98x10^24 kg

R = 6.38x10^6 m (Earth's radius)

Law Of Gravitation:

F = GmM/r^2

F = m(GM/r^2) = mg

GM/r^2 = 3.00 m/s^2

Solve for r and subtract the radius of the earth.

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