The expression E p = mgh applies only close to the surface of the Earth. The general expression for the potential energy of a mass m at a distance R from the center of the Earth (of mass m E ) is E p = − G m E m / R . Write R = R E + h , where R E is the radius of the Earth; show that, when h ≪ R E , this general expression reduces to the special case, and find an expression for g. You will need the expansion ( 1 + x ) − 1 = 1 − x + … .

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horaciocrotte horaciocrotte · Answer 1 · 2019-09-28T18:45:34+02:00

Answer:

g = G mE /Re^2

Explanation:

Hello!

First of all we need to find an equation where we can apply the series expansion of (1+x)^-1 where the variable x be a small number.

Since we are considering that h<<Re this implies that (h/Re)<<1 so this is the small number we are looking for.

Then in the equation for the earth's potential we can take of the earths radius as:

[tex]E_{p} = -\frac{G m M_e}{R} = -\frac{G m M_e}{R_{e}+h} = -\frac{G m M_e}{R_{e}}\frac{1}{(1+\frac{h}{R_{e}})}[/tex]

Now we can define x=h/Re and use the series expansion. Therefore:

[tex]E_{p} = -\frac{G m M_e}{R_{e}}(\frac{1}{1+x}) =-\frac{G m M_e}{R_{e}}(1-x+...}))[/tex]

Since x is a small number, x^2 will be even smaller, tehrefore we will only consider the first power of x:

[tex]E_{p} = -\frac{G m M_e}{R_{e}} (1-\frac{h}{R_{e}})[/tex]

and:

[tex]mgh = h\frac{G m M_e}{R_{e}^{2}}\\\\g = \frac{G M_e}{R_{e}^{2}}\\[/tex]