Newton's famous, and supposedly apple inspired, idea that the same force that caused objects to be bound to the Earth's surface was what made the planets orbit each other was a huge step forwards, pioneering the concept of forces which act a distance.

Suppose we know both the distance of the moon from the earth $r$ ($384,000 \mathrm{km}$) and the speed of it's orbit $v$ ($1022 \mathrm{ms^{-1}}$).

From our last lecture we know that the centripetal acceleration of an object is $a_{R}=\frac{v^{2}}{r}$. From the known velocity and acceleration it can be found that $a_{R}=0.00272\mathrm{ms^{-2}}$. Clearly this is much less than the acceleration due to gravity at the Earth's surface ($9.8\mathrm{ms^{-2}}$)!

So if we want to follow the idea that the same force is responsible for both the centripetal acceleration of the moon **and** the falling of an apple it is clear that the force due to gravity must depend on how far objects are from each other.

Suppose we also know the radius of the Earth ($6400 \mathrm{km}$). We can see that moon is about 60 times as far from the center of the Earth as the apple. The acceleration however is $\frac{1}{3600}$.

From this Newton concluded (although Robert Hooke also laid claim to this idea) that the dependence of the gravitational force should depend on the inverse square of the distance.

Newton's Second Law tells us that the force on the moon must be proportional to it's mass $m_{M}$.

Newton's Third Law tells us that as well as the Earth exerting a force on the Moon, the Moon should exert an equal and opposite force on the Earth, so the force needs to also be proportional to $m_{E}$.

This combination of considerations leads us to:

$F\propto\frac{m_{E}m_{M}}{r^{2}}$

We can define a Gravitational Constant $G$ such that the force between two masses is

$F=G\frac{m_{1}m_{2}}{r^{2}}$

Although Newton could determine the form of the gravitational force he could not determine the constant in his law. (He did not know the mass of the Earth or Moon.)

The gravitational constant can be determined using a torsion balance. This experiment was first performed by Henry Cavendish in the Cavendish Experiment who used it to measure the density of the Earth. Other scientists later used his results to determine the value of G.

The value of $G$ is $6.67\times10^{-11}\mathrm{Nm^{2}kg^{-2}}$.

$\vec{F}_{12}=-G\frac{m_{1}m_{2}}{r^2_{21}}\hat{r}_{21}=G\frac{m_{1}m_{2}}{r^2_{21}}\hat{r}_{12}$

$\vec{F}_{21}=-G\frac{m_{1}m_{2}}{r^2_{21}}\hat{r}_{12}=G\frac{m_{1}m_{2}}{r^2_{21}}\hat{r}_{21}$

$\vec{F}_{12}$ is the force **on** particle 1 (mass $m_{1}$) **due to** particle 2 (mass $m_{2}$).

$r_{21}$ is the distance between the two particles.

$\hat{r}_{21}$ is a unit vector which points from particle 2 toward towards particle 1.

The gravitational field $\vec{g}$ due to a mass $M$ is

$\vec{g}=-\frac{GM}{r^{2}}\hat{r}$

The normal units for gravitational field are $\mathrm{\frac{N}{kg}}$. (Note these are dimensionally equivalent to $\mathrm{ms^{-2}}$.)

The force on an object $m$ due this field is

$\vec{F}=m\vec{g}$

Looks familiar? Near the Earth's surface we can find that $\vec{g}=9.8\mathrm{ms^{-2}}$.

The form of the gravitation law we have presented implicitly assumes that we can approximate all the mass of an object as being at it's center. Now that we have an expression for the field we can show this to be explicitly true for spherical objects.

This is trivial if you know how to do surface integrals as Newton's Law of gravity can be expressed as Gauss' Law. (We'll be returning to Gauss' law in PHY 132!)

Assuming that we don't know how to do surface integrals we can also derive the result using a shell model.

We can also see from the shell model that a shell further from the center of the sphere than an object exerts **no** gravitational force on it.

What does this mean for an object that falls through the center of a planet?

An object that falls through a hole in a planet experiences a force due to all the mass that is within a sphere such that it is closer to the center of the planet than the object.

If we can assume that the density of the planet ($\rho$) is uniform then the mass within a given radius $r$ is equivalent to $\rho\frac{4\pi}{3}r^3$

The force on an object at a distance $r$ from the center is therefore

$F=Gm\frac{\rho\frac{4\pi}{3}r^3}{r^2}=Gm\rho\frac{4\pi}{3}r$

and is always directed to the center of the planet.

Whatever gains in velocity made as you fall to the center, will be lost on the way back to the surface. If the object starts at rest it will turn around at the other side and repeat the motion in the other direction, oscillating back and forward for ever. (We'll come back to this as an example of simple harmonic motion later in the course.)

The acceleration due to gravity $\vec{g}$ on the surface of any given planet of mass $M_{P}$ and radius $R_{P}$

$\large \vec{g}=-\frac{GM_{P}}{R_{P}^2}\hat{r}$

Of course it can also be useful to express this in terms of the density of the planet.

$\vec{g}=-G\frac{4\pi}{3}\rho R_{P}\hat{r}$

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

Planet | Mass | Radius |
---|---|---|

Mercury | $3.3\times10^{23} \mathrm{kg}$ | $2400 \mathrm{km}$ |

Venus | $4.87\times10^{24} \mathrm{kg}$ | $6050 \mathrm{km}$ |

Earth | $5.97\times10^{24} \mathrm{kg}$ | $6380\mathrm{km}$ |

Mars | $6.42\times10^{23} \mathrm{kg}$ | $3390 \mathrm{km}$ |

Jupiter | $1.90\times10^{27} \mathrm{kg}$ | $69900 \mathrm{km}$ |

Saturn | $5.68\times10^{26} \mathrm{kg}$ | $58200 \mathrm{km}$ |

Uranus | $8.68\times10^{25} \mathrm{kg}$ | $25400 \mathrm{km}$ |

Neptune | $1.024\times10^{26} \mathrm{kg}$ | $24600 \mathrm{km}$ |

Be careful that to find the gravitational acceleration of an object at height $h$ above the surface of the earth, the height needs to be added to the radius!

$\large \vec{g}(h)=-\frac{GM_{E}}{(R_{E}+h)^2}\hat{r}$

Also remember that this formula is not valid for $h<0$ (as we showed before) because of the excluded mass.

Useful numbers:

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

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

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

Consider a rocket launched from the surface of the earth with velocity $v_{0}$ and then subject only to gravitational force, 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

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

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.

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 field 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.

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.

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 othersâ€¦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.

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.

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.