User:Inconspicuum/Physics (A Level)/Gravitational Forces

The nature of gravity is a deep question. What exactly goes on to give rise to gravitational forces is considered one of the greatest mysteries left to solve in physics. We should all be familiar with the fact that gravity gives objects weight, which is why an apple will fall to the Earth. In fact, any object with mass exerts a gravitational force on any other object with mass. The Earth exerts a pull on an apple, and the apple equally exerts a pull on the Earth.

Gravity does not just allow us to describe the paths of objects that will eventually hit the ground. It describes how planets move, with remarkable precision. It predicts the dates and times of solar eclipses hundreds of years before they happen.

Thus gravitational force of attraction between two objects is given by:

$F_{grav}={\frac {GMm}{r^{2}}}$,

where r is the distance between the spheres, and G is the Gravitational constant. Experiments have shown that G = 6.67 x 10^-11 Nm²kg^-2.

This equation better describes some facts we already know. We should remember from GCSE science that force is inversely proportional to the distance squared. It explains why as mass increases, weight on the Earth increases.

Note that this formula says nothing about where gravity comes from. It only describes what gravity does. However, it describes it extremely well.

Gravity is not just about objects hitting the Earth. However, it does have a lot to do with falling.

We should also be aware that from circular motion that:

$F={\frac {mv^{2}}{r}}$,

where F is the centripetal force, m is the mass of an object, r is the distance between the objects and v is the velocity of the object, perpendicular to the centripetal force and thus tangential at any point to the orbit.

Gravitational Force Inside an Object

A lift acting under gravity in a lift shaft going through the centre of the Earth.

Inside a roughly spherical object (such as the Earth), it can be proved geometrically that the effects of the gravitational force resulting from all the mass outside a radius at which an object is located can be ignored, since it all cancels itself out. So, the only mass we need to consider is the mass inside the radius at which the object is located. The density of an object ρ is given by the following equation:

$\rho ={\frac {M}{V}}$,

where M is mass, and V is volume. Therefore:

$M=\rho V$

If we substitute the volume of a sphere for V:

$M={\frac {4}{3}}\pi \rho r^{3}$

And if we substitute this mass into the formula for gravitational force given above:

$F_{grav}={\frac {-Gm{\frac {4}{3}}\pi \rho r^{3}}{r^{2}}}=-{\frac {4}{3}}\pi G\rho mr$

In other words, inside a sphere of uniform mass, the gravitational force is directly proportional to the distance of an object from the centre of the sphere. Incidentally, this results in a simple harmonic oscillator such as the one on the right. This means that a graph of gravitational force against distance from the centre of a sphere with uniform density looks like this: