Grigoriy S. answered • 12/04/21

AP Physics / Math Expert Teacher With 40 Years of Proven Success

My students in AP physics 1 and C courses do this problem like this:

On the surface of the Earth the weight of the body W_{0} = mg_{0},

here m – mass of the body,

_{ }*g*_{0}* = GM / R*^{2}* = 9.8 m/s*^{2 }- acceleration due to gravity on the Earth’s surface,

where G – gravitational constant,

M – mass of the Earth.

R - radius of the planet.

If the body at a distance h above the earth surface, then its weight W = mg,

where *g = GM / (R + h)*^{2}.

In our case h = 2R, hence the formula for g looks like this:

g = GM / (3R)^{2} = GM / 9R^{2}

If we divide W by W_{0} after cancelations of mass, we get

W / W_{0} = g / g_{0}

or

W / W_{0} = R^{2} / 9R^{2} = 1 / 9

Hence, we see that weight W on this distance from the center of the Earth is 9 times smaller than on Earth's surface. Knowing that W_{0} = mg_{0}, we calculate it. W_{0} = 980 N.

Divide this number by 9 to get the weight of the body on a distance equal to 2R from the Earth’s surface. Final answer is **109 N**