Physics 221 Chapter 13

Is there gravity on Mars? Newton's Law of Universal Gravitation F = GmM/r 2 Compare with F = mg so g = GM/r 2 g depends inversely on the square of the distance g depends on the mass of the planet g on the Moon is 1 /6 of g on Earth

Example A Neutron Star’s Gravity An exotic finish to massive stars is that of a neutron star, which might have as much as five times the mass of our Sun packed into a sphere about 10 km in radius! Estimate the surface gravity on this monster.

Solution A Neutron Star’s Gravity g = GM/r 2 g = (6.67x )(5x1.99x10 30 ) / (10 4 ) 2 g = 6.7x m/s 2

Example Satellites Geosynchronous satellites are used for cable TV transmission and weather forecasting. They orbit about 36,000 km above the Earth’s surface. This is six times the radius of the Earth which is 6,000 km. What is the value of “g” at that height?

Solution Satellites The distance of the satellite from the center of the Earth is 7 R E so the acceleration due to gravity must be 1 / 49 that on the surface of the Earth. (1/49)(9.8) = 0.2 m/s 2

Kepler’s Laws 1.All planets move in elliptical orbits with the Sun at one focus 2.The radius vector drawn from the Sun to a planet sweeps out equal areas in equal intervals of time 3.The square of the orbital period of any planet is proportional to the cube of the semi-major axis of the elliptical orbit

Explanation of Kepler’s Laws (1) A planet moves faster when it is closer to the sun and slower when it is farther away from the sun (2) The angular momentum is conserved and so is the total energy (kinetic + potential) (3) GMm / r 2 = mv 2 /r ….. (1) v = 2  r / T …… (2) T 2 = (4  2 /GM) r 3 T 2 = Kr 3

Potential Energy U = mgh is not correct because g is not constant! Need to integrate … ain’t that fun ;-) U f – U i = GMm ∫ 1/r 2 dr U f – U i = - GMm [ 1/r f - 1/r i ] Note: U i = 0 when r i =  U r = - GMm / r

Energy in Planetary Motion E = K.E. + P.E. E = ½ mv 2 – GMm/r Total energy is conserved (constant). If K.E. increases P.E. decreases.

Escape Speed What is the minimum speed of launch required to escape the Earth’s gravity? ½ mv 2 = GMm/R v esc = [2 G M / R] 1/2

Black Holes The escape speed exceeds the speed of light! The critical radius where the escape speed is just equal to the speed of light c is called the Schwarzchild radius which defines the event horizon.

That’s all folks!