To answer your question, let's just place the formula F = G * m1 m2 / R^2 into a Taylor series expansion (as distance from the center of he earth is changed) and compare it with the approximation F = mg.
So, at Earth's surface
F = G m1 m2 / R^2
But if we to go just a little higher (say a height of h) than Earth's surface, we would have
F = G m1 m2 / (R+h)2
F = G m1 m2 / (R2
F = G m1 m2 / R2
Now, take note that since F = mg is an approximation for the above formula at h << R, we can take a Taylor series expansion of (1+h/R)-2
This expansion would be 1 - 2h/R + 3(h/R)2
+-... and so on.
Since h/R is close to 0 when h << R, we can just take the first term "1" of the Taylor series expansion to approximate F = G m1 m2 / R^2. From this expansion, we can also see that unless h is big enough to cause the next few terms "-2h/R" and "3(h/R)2
" to be significant, F = mg is a pretty good approximation for the gravitational force at heights close to the Earth's radius.
I'd also like to show that the two formulas are numerically identical.
Formula 1: F = m * (g)
where F is the amount of force felt by an object of mass m, and g is the acceleration due to Earth's gravity near its surface.
Formula 2: F = m1
* (G * m2 / r2
where F is the amount of force felt by an object of mass m1 due to an object with mass m2
at a distance of r apart from each other. G is the gravitational constant.
So, the numerical value of g is commonly used as 10 ms-2
, or more accurately 9.81 ms-2
G is gravitational constant, and its value is 6.67300 * 10-11
is actually the mass of the Earth, and its value is 5.9742 * 1024
r is actually the radius of the Earth, and its value is 6.3781 * 106
Now, if you calculate the value of the term (G*m2/r^2), we get a value that is 9.7998 ms-2
-- which is extremely close to the value of g (9.8ms-2
Hope that clears things up!
(published on 05/17/12)