# About the work done by gravity to pull the object down to the earth surface

The following question is from an physics exercise and I know the answer and the way to solve the problem but just curious why my own way doesn't work. The question is asking "A 2500kg space vehicle, initially at rest, falls vertically from a height of 3500km above the Earth's surface. Determine how much work is done by the force of gravity in bringing the vehicle to the Earth's surface." The way to solve that is to integrate the differential work from the force $F_g = GM_\text{earth}m/r^2$ such that $$W = \int_{6400 \times 10^3}^{(6400+3500)\times 10^3}F_g dr$$. It does solve the problem but my question is why can't I use the $mgh$ to estimate the work done by gravity?

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Because $mgh$ is not the same as $GMm/r^2$. $mgh$ is an approximation to the true force ($GMm/r^2$) that applies when $h\ll r$. This is okay for ordinary day to day situations since you never experience things more than a few km up. But it breaks as soon as you go farther out. – Michael Brown Mar 28 '13 at 3:07
I see. I just don't know it is an approximation. Thanks. – user1285419 Mar 28 '13 at 3:18
I have one more question. I was told in the class that gravity is conservative force. But if we consider the general case with $GMm/r^2$, is that statement still true? – user1285419 Mar 28 '13 at 3:26
Yes, still true. – Michael Brown Mar 28 '13 at 3:44
user1285419 - it is still conservative because the total energy of the system (i.e. the space-craft/Earth system) never changes. This of course assumes an ideal system with no friction. The expression of $GMm/r^2$ just happens to be the expression that describes how the force of gravity changes in our observed universe... it is an experimental fact. In another universe that expression might be different, but in that other universe if the force was conservative, then regardless of the mathematical expression of the force, the total energy of the system (potential + kinetic) would be constant. – davecoulter Mar 28 '13 at 5:30

Your use of "$mgh$ to estimate" is spot on. You can see in the $GMm/r^2$ formula that the force between the bodies differs with their distance, squared. The $mgh$ formula assumes $g$ is constant, independent of $h$ which obviously isn't true in $GMm/r^2$. As the object falls the radius shrinks and the value used for $g$ actually changes.

3500km is pretty high up compared to the radius of the earth considering $g$ is normally taken at the surface.

By integrating you're accounting for the change in the force over the whole 3500km distance.

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Remember that the force of gravity is $F_g = \frac{-GMm}{r^2}$. But, if $r$~constant=radius of earth (i.e. the object doesn't move much, we can write $F_g = -mg$ where $g=\frac{GM}{r^2_{earth}}$.

In this case then, noting that $F_g = -\frac{d}{dy}U_g$ where $U_g$ is the potential energy, we find that $U_g = mgy$.

However, this approximation fails for large changes in $r$ (i.e. $r$ is not constant and not equal to the radius of earth). In that case we have to go back to $F_g = \frac{-GMm}{r^2}$ and thus use the potential energy $U_g = \frac{-GMm}{r}$

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