Find the altitude above the Earth's surface where Earth's gravitational field strength would be three-fourths of its value at the surface. Assume re = 6.371 multiplied by 103 km.
km
Find the altitude above the Earth's surface where Earth's gravitational field strength would be one-fourth of its value at the surface.
km
km
Find the altitude above the Earth's surface where Earth's gravitational field strength would be one-fourth of its value at the surface.
km
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R¾ = √(4/3)*Re = 7.3566E3 km; H¾ = .9856E3 km
R¼ = √4*Re = 12.742E3 km; H½ = 6.371E3 km
R¼ = √4*Re = 12.742E3 km; H½ = 6.371E3 km
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F is proportional to 1/r^2
If you want F to be 0.75 of the surface value, r needs to be 1/sqrt(0.75) of the radius at the surface.
If you want F to be 0.25 of the surface value, r needs to be 1/sqrt(0.25) of the radius at the surface.
r is the distance from the center. I'll let you figure out how to go from that to altitude (distance from the surface).
If you want F to be 0.75 of the surface value, r needs to be 1/sqrt(0.75) of the radius at the surface.
If you want F to be 0.25 of the surface value, r needs to be 1/sqrt(0.25) of the radius at the surface.
r is the distance from the center. I'll let you figure out how to go from that to altitude (distance from the surface).