Is mgh Always Applicable for Calculating Gravitational Potential Energy?

[BlackWyvern]

I don't have an exact question here. But for example, there is a 10kg mass at a distance of 2R from the centre of mass of the Earth. It's gravitational potential energy is equal (or so it's been preached by teachers) to mgh. The local gravity at 2R is approximately 2.5ms^-2. So let's say the object falls down a little bit. The height has decreased by a certain amount, and the local gravity at this point is higher. I'm pretty sure that the increase in local gravity means that the Ep worked out at 2R is wrong.

I'm pretty sure that mgh ONLY applies when the h is small enough that the g doesn't change. That it's meant for a UNIFORM acceleration. The thing is, gravity isn't a uniform acceleration.

So how do we find the true potential energy of an object in a gravitational field?

If by some virtue, you can prove that mgh IS correct for gravitational fields (which I HIGHLY HIGHLY HIGHLY doubt), then how can you find the Ep of a mass in a uniform acceleration field?

 

[robphy]

http://hyperphysics.phy-astr.gsu.edu/hbase/gpot.html

at a distance $$r \geq R_{earth}$$ from the center of the earth
$$U=-\frac{GM_{earth}m}{r}$$

just above the Earth's surface, $$r=R_{earth}+h$$, where $$0 \leq h \ll R_{earth}$$,
$$U=-\frac{GM_{earth}m}{R_{earth}+h}=-m\frac{GM_{earth}}{R_{earth}+h}=<br /> -m\frac{GM_{earth}}{R_{earth}} {\color{red}\frac{1}{(1+\frac{h}{R_{earth}} )}}$$
Can you use the Taylor series on the expression in red?
then compute the difference in potential energy from that on the Earth's surface?

 

[BlackWyvern]

Thanks very much. :)

I'm pleased that I wasn't wrong, lol.

Also, Ep = mah would be applicable for any distance (until the speed is a large percent of c) in a field where the acceleration is uniform? i.e. on a theoretical plane where gravity is constant?