Q:

I was wondering how can I show that "F=mg" is actually an approximation of "F=Gm1m2/r^2" using Taylor Series...
I'm not sure if mass is the dependent variable or radius?

- Farahnaz (age 18)

Toronto

- Farahnaz (age 18)

Toronto

A:

Hello Farahnaz,

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 / (R^{2} (1+h/R)^{2})

F = G m1 m2 / R^{2} * (1+h/R)^{-2}

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} - 4(h/R)^{3} +-... 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 = m_{1} * (G * m2 / r^{2})

where F is the amount of force felt by an object of mass m1 due to an object with mass m_{2} 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} m^{3}kg^{-1}s^{-2}.

m_{2} is actually the mass of the Earth, and its value is 5.9742 * 10^{24} kg.

r is actually the radius of the Earth, and its value is 6.3781 * 10^{6} m.

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!

Maurice

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)

F = G m1 m2 / (R

F = G m1 m2 / R

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)

This expansion would be 1 - 2h/R + 3(h/R)

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)

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 = m

where F is the amount of force felt by an object of mass m1 due to an object with mass m

So, the numerical value of g is commonly used as 10 ms

G is gravitational constant, and its value is 6.67300 * 10

m

r is actually the radius of the Earth, and its value is 6.3781 * 10

Now, if you calculate the value of the term (G*m2/r^2), we get a value that is 9.7998 ms

Hope that clears things up!

Maurice

*(published on 04/14/2012)*