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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 + … .

User Chamika
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1 Answer

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Answer:

g = G mE /Re^2

Step-by-step 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:


E_(p) = -(G m M_e)/(R) = -(G m M_e)/(R_(e)+h) = -(G m M_e)/(R_(e))(1)/((1+(h)/(R_(e))))

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


E_(p) = -(G m M_e)/(R_(e))((1)/(1+x)) =-(G m M_e)/(R_(e))(1-x+...}))

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


E_(p) = -(G m M_e)/(R_(e)) (1-(h)/(R_(e)))

and:


mgh = h(G m M_e)/(R_(e)^(2))\\\\g = (G M_e)/(R_(e)^(2))\\

User Dineen
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