Question #bfe12

1 Answer
Jan 5, 2018

The formulae are #F=(Gm_1m_2)/r^2# and #E=-(Gm_1m_2)/r# (minus because the zero is at #r=\infty#), into which we substitue #g# to remove the unknowns.

Explanation:

Since the force on a mass #m# at the Earth's surface is #mg# and also (from the formula above) #(GMm)/R^2# (where #M# is the mass of the Earth), we have

#g=(GM)/R^2#, from which we derive #GM=gR^2#.

So the potential energy may be written as #E=-(mgR^2)/r#.

For a mass at height #R/4# above the Earth's surface #r=R+R/4=5/4R#, which, inserted into the formula for #E#, gives
#E=-4/5mgR#.
Note that this is with respect to the potential at #r=\infty#; what we (presumably) want is the potential with respect to that at the Earth's surface, which should then be simply subtracted:
#E=-4/5mgR - (-mgR)=1/5mgR#.

Note that had we assumed the force to be constant with #r#, which it is not, we would have got #E=1/4mgR#, which is wrong.