Consider a rocket launched from the surface of the earth with velocity $v_{0}$, if it's height above the ground is $x$ then

$\frac{dv}{dt}=\frac{-GM_{E}}{(R_{E}+x)^{2}}$

$\frac{dv}{dx}\frac{dx}{dt}=v\frac{dv}{dx}=\frac{-GM_{E}}{(R_{E}+x)^{2}}$

$\int^{v}_{v_{0}}v\,dv=\int^{x}_{0}\frac{-GM_{E}}{(R_{E}+x)^{2}}\,dx$

$\frac{1}{2}(v^{2}-v_{0}^2)=\frac{GM_{E}}{R_{E}+x}-\frac{GM_{E}}{R_{E}}$

$v^{2}=v_{0}^2+2(\frac{GM_{E}}{R_{E}+x}-\frac{GM_{E}}{R_{E}})$

We can see whether or not the rocket achieves escape velocity by considering the limit as $x\to\infty$

$v^{2}=v_{0}^2+2(\frac{GM_{E}}{R_{E}+x}-\frac{GM_{E}}{R_{E}})$

as $x\to\infty$ the object will escape if

$0\geq v_{0}^{2}-\frac{2GM_{E}}{R_{E}}$

For the object to escape

$v_{0}\geq \sqrt{\frac{2GM_{E}}{R_{E}}}$

$M_{E}=5.98\times10^{24}\mathrm{kg}$

$R_{E}=6380\mathrm{km}$

$G=6.67\times10^{-11}\mathrm{Nm^{2}kg^{-2}}$

$\to v_{0}=11,200\mathrm{ms^{-1}}$

For a circular orbit a satellite of mass $m$ must have a velocity such that the gravitational acceleration is the same as the centripetal acceleration.

$G\frac{mM_{E}}{r^{2}}=\frac{mv^{2}}{r}$

$v_{circular}=\sqrt{\frac{GM_{E}}{r}}$

Cick here for the original link to this video from Mike Zingale.

When we did this lecture two years ago the Upper Atmosphere Research Satellite was coming down!

Here are links for the dramatic take on the event.

For a satellite to have a fixed position above the Earth's surface it must have the same **angular** velocity as a point on the Earth's surface. Another way of looking at this is that the satellite must take exactly one day to complete it's path.

$\frac{2\pi r}{v}$ = 1 day = 86,400 s

Using the equation for the velocity $v=\sqrt{\frac{GM_{E}}{r}}$

$2\pi\frac{r^{3/2}}{\sqrt{GM_{E}}}$ = 86,400 s

$r=(86,400\frac{\sqrt{GM_{E}}}{2\pi})^{2/3}=4.23\times10^{7}\,\mathrm{m}=42,300\,\mathrm{km}$

(36,000 km above surface)

Mass of Earth $M_{E}=5.9742\times10^{24}\,\mathrm{kg}$

Gravitational Constant $G=6.67\times10^{-11}\,\mathrm{Nm^{2}kg{-2}}$

GPS satellites are essentially extremely accurate clocks that continually broadcast their position and the time, which an Earth based receiver can use to calculate it's own position, so long as it can receive at least 4 signals. It is not desirable for GPS satellites to be in geostationary orbits.

A GPS satellite orbits about 20,200 km above the surface of the Earth (or at a radius of 26,600 km from the center of the Earth.)

How many times a day do the GPS Satellites go round the Earth?

$v=\sqrt{\frac{GM_{E}}{r}}$

Mass of Earth $M_{E}=5.9742\times10^{24}\,\mathrm{kg}$

Gravitational Constant $G=6.67\times10^{-11}\,\mathrm{Nm^{2}kg^{-2}}$

Number of seconds for a GPS satellite to go round the Earth $\frac{2\pi r}{v}=2\pi\frac{r^{3/2}}{\sqrt{GM_{E}}}=2\pi\frac{(2.66\times10^{7})^{3/2}}{\sqrt{6.67\times10^{-11}\times5.9742\times10^{24}}}=43,200$

Times round the earth in a day = Seconds in one day/Number of seconds for a satellite to go round the Earth

Seconds in one day = $24\times60\times60=86400$

Times round the earth in one day $\approx 2$

$v=3870\mathrm{m/s}$

This is fast enough that the clock on a GPS satellite is slightly slower than one on Earth because of special relativity. However due to general relativity the difference in the gravitational feed between the surface of the Earth and the height of the satellite causes the clock to go faster than on one Earth. These two effects need to be taken in to account for the GPS system to work.

Recall from the last lecture that the escape velocity is

$v_{E}=\sqrt{\frac{2GM_{E}}{R_{E}}}$

Compare this to

$v=\sqrt{\frac{GM_{E}}{r}}$

and $r>R_{E}$

So there are velocities above the velocity required for a circular velocity, but less than escape velocity. These lead to elliptical orbits.

Cick here for the original link to this video from Mike Zingale.

The elliptical orbits of the planets can be described by Kepler's Laws. These were advanced nearly a century before Newton's Law of Gravitation by Johannes Kepler.

- The orbit of every planet is an ellipse with the Sun at a focus.
- The line joining a planet and the Sun sweeps out equal areas during equal intervals of time.
- The square of the orbital period of a planet is directly proportional to the cube of the semi-major axis of it's orbit

An interesting article about Kepler and Tycho Brahe in Physics Today.

The orbit of every planet is an ellipse with the Sun at a focus.

$F_{1}P+F_{2}P=2s$

$e$ is the eccentricity and $0<e<1$.

The eccentricity is the deviation of the orbit from circular, Earth's eccentricity is 0.017, which is not very much.

Cick here for the original link to this video from Mike Zingale.

The orbits of most of the planets have low eccentricity. Pluto is more eccentric than the others…but it is also no longer a planet, mainly because we have found many more objects out there which are better described as dwarf planets. Sedna is the most distant known object in the solar system andis about two-thirds the size of Pluto..and as you can see has a very long and eccentric orbit.

The line joining a planet and the Sun sweeps out equal areas during equal intervals of time.

This means that planets move faster when they are closer to the sun.

Cick here for the original link to this video from Mike Zingale.

Comet orbits can have much higher eccentricity than planets.

Orbits of the Kohoutek Comet (red) and the Earth (blue), illustrating the high eccentricity of its orbit and its rapid motion when close to the Sun.

The square of the orbital period of a planet is directly proportional to the cube of the semi-major axis of it's orbit

So if two planets have semimajor axes $s_{1}$ and $s_{2}$ then their periods can be related to each other via the relation

$\large (\frac{T_{1}}{T_{2}})^{2}=(\frac{s_{1}}{s_{2}})^{3}$

It follows that $\frac{s^{3}}{T^{2}}$ should be the same for each planet.

Cick here for the original link to this video from Mike Zingale.

We will derive Kepler's Third Law for the special case of a circular orbit.

We consider only the force between the sun and the planet (ie. we ignore the force from the other planets.)

$G\frac{m_{p}M_{s}}{r_{p}^{2}}=m_{p}\frac{v_{p}^2}{r_{p}}$

$\large v_{p}=\frac{2\pi r_{p}}{T_{p}}$

$G\frac{m_{p}M_{s}}{r_{p}^{2}}=m_{p}\frac{4\pi^2 r_{p}}{T_{p}^2}$

$\large \frac{T_{p}^2}{r_{p}^3}=\frac{4\pi^{2}}{GM_{s}}$

$T_{p}^2=\frac{4\pi^{2}}{GM_{s}}r_{p}^3$

So for a circular orbit we have shown that the square of the orbital period of a planet is directly proportional to the cube of the semi-major axis of it's orbit ($r_{p}$ for a circle).

The Sun's gravitational pull on the Moon is actually more than twice that of the Earth, and it's path around the Sun is not all that different to the Earth. However, its orbit around the Earth, is subject to several pertubations.