Gravitation Part III Orbits 1 The Importance of Orbits Examples: Launching satellites and probes, exploring space Can use to measure masses (e.g., most fundamental way of measuring masses of stars, their most fundamental property. Also masses of extrasolar planets). Clues to origin and history of whatever is orbiting (e.g. comet orbits imply huge reservoirs of unseen comets, orbits of stars and gas in galaxies imply existence of Dark Matter).

2 Gravity's predictive power Halley's comet: Edmund Halley found orbit of comet seen in 1682 appeared similar to other comets spotted 1607 and 1537 => predicted it would return in 1758/59. It did! The discovery of Neptune: 1781 Herschel discovered Uranus, but by 1840 the predicted positions were clearly disagreeing with measurements => gravitational influence from another body. Thus eighth planet predicted and found 1845! 3

Reminder: Newtons generalized form of Kepler's 3rd law 2 3 4 2 P a G (m1 m2 ) a is mean separation of the two objects over the orbit. The solar system is a special case: if m1 is the mass of the Sun, it is much larger than the mass of any planet (even Jupiter is 1000x less massive), so 4

m1+m2 ~ Msun. Binary stars 5 More examples: determine the total mass of a system The Centauri system is 1.338 pc distant with a period of 79.92 years. The A and B components have a mean separation of 23.7 AU (although the orbits are highly elliptical). What is the total mass of the system? First way: use the generalized form of Kepler's third law in mks units. 2

3 4 2 P a G (m1 m2 ) 6 P 79.92 yr 365 days/yr 24 hr/day 3600 s/hr 2.52 109 s a 23.7 AU 1.496 1011 m/AU 3.55 1012 m 4 2 a 3 4 2 (3.55 1012 m) 3 30 (m1 m2 )

4 . 2 10 kg 2 11 3 2 9 2 GP 6.67 10 m /kg/s (2.52 10 s)

Unit check: m3/(m3 kg-1 s-2 s2)=kg Compare to mass of our Sun = 1.99x1030 kg 3 Second way: use same law but written as P 2 a m1 m2 as long we use correct units (years, AU, 7 solar masses). Another example: geosynchronous satellites At what height above the surface of the Earth are geosynchronous satellites located? Geosynchronous orbits are defined to keep the satellite above the same place on Earth, which means they have a period

P=23.93 hrs (why not 24 hrs?). a is mean separation of the objects. For circular orbits where one is much more massive, a is approximately the orbit radius of the less massive one. 8 P 23.93 hr/day 3600 s/hr 8.6110 4 s m1 m2 mEarth 5.97 1024 kg GP 2 mEarth 6.67 10 11 m 3 /kg/s2 (8.61104 s) 2 (5.97 1024 kg) 22 3 a 7 .

48 10 m 4 2 4 2 a 4.21107 m 42,100 km 3 Now subtract off Earths radius to get the altitude: 35,750 km. Can you have a geosynchronous satellite anywhere else than above the equator? 9 Weight versus mass - reminder

Mass is a measure of how much matter an object has Weight is a measure of how strong gravity pulls on the object Weight is felt by the support given to us by the floor, seat, etc. The mass stays the same no matter what force acts upon an object Is your mass greater or smaller on the Moon than on the Earth? 10 Weightlessness Astronauts on the space station feel weightless because: a) there is no gravity in space and they do not weigh anything.

b) space is a vacuum and there is no gravity in a vacuum. c) the astronauts are far from Earth's surface at a location where gravitation has a minimal affect. 11 Weightlessness Astronauts on the space shuttle feel weightless because: a)

there is no gravity in space and they do not weigh anything. but then what would make them orbit in the first place? b) space is a vacuum and there is no gravity in a vacuum. not true c) the astronauts are far from Earth's surface at a location where gravitation has a minimal affect. At 400 km orbit, g is reduced from 9.8 to 8.7 m/s22 - not enough 12

Weightlessness Weightlessness happens when nothing supports you against gravity In orbit no force counteracts gravity - you are constantly accelerating. In "free-fall". So is the space station and everything in it. So it does not push up on you like the Earth does => simulates gravity-free environment 13 So if gravity is the force causing the revolution of the Moon around the Earth, and the planets around the

Sun, why doesnt the Moon fall straight into the Earth or the planets plunge into the Sun? Thought experiment: imagine throwing a rock from a great height above the Earths surface. 14 Paths A-F correspond to harder and harder throws. If hard enough, the rock simply misses the Earth altogether. Cases D-F are orbits of the rock around the Earth. The type of orbit depends on the injection velocity.

15 Example: How fast do you need to throw rock to get it into circular orbit? For circular motion, for a small mass, the acceleration is the centripetal acceleration: acentripetal So since then Vcircular r

2 Fgravity macentripetal GM Earth m rock Vcircular m rock 2 r r 2 16 Canceling mrock and one factor of 1/r,

GM Earth 2 Vcircular r or Vcircular GM Earth r This circular speed is the magnitude of the injection

velocity (in direction perpendicular to center of Earth) needed for a circular orbit at distance r from the Earth. 17 This is general! Example: what is the circular speed at 1 AU from the Sun for a small mass? M = MSun = 2.0 x 1030 kg; r is 1.5 x 1011 m, and the constant G is 6.7 x 10-11 in SI units, so Vcircular (6.7 10 11

3 2 30 m /kg/s )(2.0 10 kg ) 11 1.5 10 m 4 Vcircular 3.0 10 m/s Vcircular 30 km/s Accurate average speed of the Earth around the Sun:

29.79 km/s 18 Keplerian rotation curves Consider circular orbits for simplicity. When the system is dominated by the central mass: v r -1/2 19 Milky Way also rotates. But rotation curve not Keplerian, but nearly flat. Milky Way mass is not dominated by a central mass 20 What if injection velocity isnt the circular velocity?

In general, orbits are conic sections (Newton). Ellipses and circles are closed orbits. Hyperbolic and parabolic orbits are open objects on these orbits do not return (some comets do this!). 21 Escape speed How large must the injection speed be to make the object escape gravity of the body it was launched from altogether? Quantitatively, this means object gets infinitely far away from the body it is escaping when its speed has dropped to zero (if speed = 0 km/s at finite distance, gravity would bring it back) Examine in terms of energies. Its kinetic energy is:

At escape, then, KE = 0. 1 2 KE mV 2 22 When it escapes, what is its potential energy? Potential energy is energy due to its position, in the presence of a force. GMm PE

r At escape, its PE is also 0. Its total energy is then TE = KE + PE, which is zero when it escapes. 23 Total energy is conserved, therefore it is also 0 at launch: GMm 1 2 mVescape 0 r 2 Vescape

2GM r Escape speed (see box 7.2 in text) Notice this is bigger than circular speed. Makes sense: you are not escaping if going at circular speed. This relation is very useful! Example: Vescape from the surface of the Earth is 11.2 km/ 24 s (check for yourself).

Another example A neutron star is a dead star with about twice the mass of the Sun, but a radius of only 10 kilometers. How fast would you have to be going to escape the surface of a neutron star? 2GM 2 6.67 10 11 m 3s -2 kg -1 4 1030 kg Vescape r 10 4 m 2.3 108 m/s 25 General rules of orbits

Orbit shape Total energy Parabolic Zero Elliptical/circular Negative Hyperbolic Positive

Negative energy orbits are bound. Positive ones unbound (think of a fly-by). 26 27 Newton's three laws of motion 1. A body remains at rest, or moves in a straight line at a constant speed, unless acted upon by a net outside force. 2. The acceleration of an object is proportional to the net outside force acting on the object, and inversely proportional to its mass (F=ma). 3. Whenever one body exerts a force on a second body, the second body exerts an equal and opposite force on

the first body. 28