Physics 221
Chapter 13
Is there gravity on Mars?
Newton's Law of Universal Gravitation
F = GmM/r2
Compare with F = mg so
g = GM/r2
• 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 13.1 . . . 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 13.1 . . . A Neutron Star’s Gravity
g = GM/r2
g = (6.67x10-11)(5x1.99x1030) / (104)2
g = 6.7x 1012 m/s2
Example 13.2 . . . 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 13.2 . . . Satellites
The distance of the satellite from the center
of the Earth is 7 RE so the acceleration due
to gravity must be 1 / 49 that on the surface
of the Earth.
(1/49)(9.8) = 0.2 m/s2
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 / r2 = mv2 /r ….. (1)
v = 2  r / T ……...........(2)
T2 = (4 2 /GM) r3
T2 = Kr3
Potential Energy
U = mgh is not correct because g is not constant!
Need to integrate … ain’t that fun ;-)
Uf – Ui = GMm ∫ 1/r2 dr
Uf – Ui = - GMm [ 1/rf - 1/ri ]
Note: Ui = 0 when ri = 
Ur = - GMm / r
Energy in Planetary Motion
E = K.E. + P.E.
E = ½ mv2 – 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 mv2 = GMm/R
vesc = [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!