in the equation,g'=g<1-[2h/R]>;WHY CAN'T h=infinity

in the equation,g'=g<1-[2h/R]>;WHY CAN'T h=infinity

 

Where does the equation come from? i.e. it's probably derived somehow from g = GM/R^2, so what assumptions are made in simplifying the formula?

actually it is the formula of variation of g with altitude,mr.swansot

actually it is the formula of variation of g with altitude,mr.swansot

 

But where is it from? I have tried to derive it - and I can only get there if I ignore higher orders of h/R and maddeningly drop a value of 2Rh. This might make sense when h is incredibly small compared to R - but you are positing "what if h is infinite!". In that case of h being large or even of the same order as R you need to rework the equation without the ignoring of orders of h/R and massive simplifications.

 

In reality it is an approximation that works for human scale structure on the earth - but not for large scale changes in height.

actually it is the formula of variation of g with altitude,mr.swansot

 

Variation with altitude is a specific application of a more general equation, in this case, Newton's law of gravitation. As imatfaal has said, it has to come from somewhere. When you have an equation in physics, you need to know what assumptions went into the derivation, because that has implications on the limits of applicability of the equation.

 

You can derive your equation by looking at the ratio of two values of g at a separation of h, and do a binomial expansion. But (again, as imatfaal has said) to get that particular equation, you have to make an assumption about the value of h/R in order to truncate the series and ignore the higher-order terms — you have to assume that h << R.

I believe the derivation of that equation probably went something like this: start with the g-value on Earth's surface:

 

[math]g=\frac{GM}{R^2}[/math]

 

Now define g' as the g-value at a height h above the Earth's surface:

 

[math]g'=\frac{GM}{(R+h)^2}=\frac{GM}{R^2} \left (1+\frac{h^2}{R^2}+\frac{2h}{R} \right )^{-1}=g \left (1+\frac{h^2}{R^2}+\frac{2h}{R} \right )^{-1}[/math]

 

Now do a binomial expansion to get:

 

[math]g'=g \left (1-\left (\frac{h^2}{R^2}+\frac{2h}{R} \right )+\left ( \frac{h^2}{R^2}+\frac{2h}{R} \right )^2 - \left ( \frac{h^2}{R^2}+\frac{2h}{R} \right )^3 + ... \right )[/math]

 

Now if we make the assumption that h<<R, (h is very small compared to R) then terms with higher powers of (h/R) are essentially zero. This makes all terms in our expansion go to zero, except:

 

[math]g'\approx g \left (1-\frac{2h}{R} \right )[/math]

 

If you make h large compared to R, then the simplifying assumption we used no longer holds. In other words, the approximation fails when h becomes large. So h=infinity is definitely not going to work.

You can do the binomial expansion before squaring. It's a bit simpler. (1+h/R)^-2 expands directly to (1-2h/R)