# Gravitational Questions

I come across this formula :
g-g'=(2h/R)g ; h<R
However , if I substitute h =3R/4 the change in gravitational will be 1.5g .
How can the change of gravitational field strength is bigger than its original gravitaitonal field strength ?

Besides , how can I prove F=GMm/R2 ?

Besides this , where can I find the details explanation on gravitational

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quasar987
Homework Helper
Gold Member
?! Where did you find this formula?

P.S. You can't prove F = -GMm/r². Newton defined force as F = ma and then calculated that a grav. force of the form F = -GMm/r² would explain both the motion of the planets and the falling of apples at the surface of earth. From there, we can only make observations that will either agree with his theory of gravitation, or disprove it. But never prove it.

Edit: What I meant to say is that F = -GMm/r² is like THE axiom of Newton's theory of gravitation. Historically, this was the mother equation, and all the other flowed from it. Of course, starting from different axioms, you could manage to derive, and thus "prove" F = -GMm/r². But it would only be a proof in the mathematical sense of the word, meaning that F = -GMm/r² is true provided your axioms are true. But since no axioms on which a physical theory is built are absolutely provable, nothing is absolutely provable, including F =-GMm/r².

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lightgrav
Homework Helper
The local difference in gravity strength at different heights
(g_lo - g_hi) is only APPROXIMATELY equal to 2g(h/R) , and
only if h be MUCH smaller than R : h << R .

We START with g = GM/R^2 . If we move to a higher elevation,
R becomes $$R_{hi} = R_{lo} + h$$ ; $${R_{hi}}^{2} = {R_{lo}}^{2} + 2 h R_{lo} + h^{2}$$ .
We ignore h^2 compared to R.
Now, the fractional "%" difference $$({R_{lo}}^2 - {R_{hi}}^2) / {R_{lo}}^2 \approx 1 - 2h/R \approx \frac{1}{1+2h/R}$$
So g_old , at low elevation , changes by the reciprocal of that factor,
$$\frac{g_{lo}-g_{hi}}{g} \approx 1 + 2h/R$$ .

(this is easier with derivatives)

You can "prove" that nested sperical shells (sort-of-like Earth)
are surrounded by a gravity field "as if" they were all at R = 0 ...

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