Asked by johnny
At what altitude above Earth's surface would the gravitational acceleration be 3.0 m/s2?
Answers
Answered by
Damon
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
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
Answered by
GK
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.
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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