Newton's Law of Gravitation is given by:
$F = -G \frac{Mm}{r^2}$
Where M and m are the masses of the two objects, r is the distance between them, and G is the gravitational constant.
You can express it as a vector (in fact if you are simulating motion in a gravitational field you need to express it as a vector):
$\textbf{F}= -G \frac{Mm}{r^2}\hat{\textbf{r}}$
where $\hat{\textbf{r}}$ is the unit vector in the direction from the mass M to m.
Gravitational Potential Energy
Gravitational potential energy is defined as the work done to bring an object from a distance r to infinity.
$U = \int^\infty_r F dr $
$U = \int^\infty_r \frac{-GMm}{r^2} dr$
$U = -GMm \int^\infty_r r^{-2} dr$
$U = -GMm [-r^{-1}]^\infty_r$
$U = GMm \left[\frac{1}{r}\right]^\infty_r$
$U = GMm [0 - 1/r]$
$U = -\frac{GMm}{r}$
Conservation of Energy
An object of mass m moving with velocity $v_1$ at distance $r_1$ from a massive body of mass M has the combined gravitational potential energy U and kinetic energy K:
$U + K =-G\frac{Mm}{r_1} + \frac{1}{2}mv_1^2$
If no energy is added into the system by using an engine, this total is conserved. At distance $r_2$ and velocity $v_2$ we will find the total is the same.
$-G\frac{Mm}{r_1} + \frac{1}{2}mv_1^2 = -G\frac{Mm}{r_2} + \frac{1}{2}mv_2^2$
If we know the speed of an object at one distance, we can predict the velocity when the object is at a different distance.
Escape Velocity
Imagine you are standing on the surface of the Earth, and tossed a ball up. You will see the ball gradually slow down, briefly stopping in the air, and then come down again. If you throw the ball hard enough, the ball will slow down, and it will stop at infinity! This means the ball has completely escaped. The velocity that you throw the ball so that it escapes is called the escape velocity.
When you are standing on the Earth, the surface is about 6371 km from the centre.
$r_1 = 6.37 \times 10^6 m$
When the ball reaches infinity, the velocity is zero:
$r_2 =\infty$
$v_2 = 0$
We want to find the escape velocity, $v_1$.
Using the principle of conservation of energy:
$-\frac{GMm}{r_1} + \frac{1}{2} mv_1^2 = \frac{GMm}{\infty} + \frac{1}{2}m (0)^2 $
$ \frac{1}{2} mv_1^2 = \frac{GMm}{r_1}$
$ \frac{1}{2} v_1^2 = \frac{GM}{r_1}$
$ v_1^2 = 2\frac{GM}{r_1}$
$ v_1 = \sqrt{2\frac{GM}{r_1}}$
Putting in the numbers, with $G = 6.67 \times 10^{-11} m^3 kg^{-1} s^{-2}$ and mass of Earth $M=5.97\times10^{24}$:
$v_1 = \sqrt{\frac{2 \times 6.67\times 10^{-11} \times 5.97\times 10^{24} }{ 6.37\times 10^6}} = 1.09\times 10^4 ms^{-1}$
That is 39,240 km/h.
What if you are already in space, at some distance away from the Earth? Let's say at a distance where $r_1 = 20000 km$? I invite you to try calculating it.
You can also try to calculate the escape velocity from the surface of other planets such as:
| Planet |
mass |
radius |
| Mars |
$6.39 \times 10^{24} kg$ |
3389.5 km |
| Jupiter |
$1.898 \times 10^{27} kg$ |
69911 km |
| Asteroid Ceres |
$9.38 \times 10^{20}$ kg |
470 km |
