Chapter 12
Gravity
DEHS 2011-12
Physics 1
Newton’s Law of Universal Gravity
• The force of gravity between any two point objects
of mass m1 and m2 is attractive and of magnitude
F G
m1m 2
r
2
• Here, r is the distance between the masses
• G is called the universal gravitation constant and it
is equal to
11
2
2
G  6.67  10
N  m kg

Inverse Square Dependence &
Superposition
• Newton’s Law of Universal
Gravitation (LUG) is an
example of an inverse square
dependence
– The strength of gravity falls off
rapidly with distance, but it
never becomes zero
– Gravity has an infinite range
• The net gravitational force like
other field forces can be found
using the principle of
superposition, it is the vector
sum of each other the forces
individually
Example 12-1
An astronaut is a distance d from his spaceship when he
experiences a gravitational pull of 80 N from his ship. He
drifts out to a distance of 4d. Now what gravitational force
does the ship exert on the astronaut?
Example 12-2
The picture shows an arrangement of three particles, particle 1 of mass
m1 = 6.0 kg, particles 2 and 3 of mass m2 = m3 = 4.0 kg, and a distance
a = 2.0 cm. What is the net gravitational force on particle 1 due to the
other particles?
Example 12-3
The picture shows an arrangement of five particles, with masses
m1 = 8.0 kg, m2 = m3 = m4 = m5 = 2.0 kg, and with a = 2.0 cm
and θ = 30°
Gravity due to a Sphere
The net force exerted
by a sphere whose
mass is SPHERICALLY
distributed (does not
have to by uniformly
distributed)
throughout an object
is the same as if all
of the sphere’s mass
were concentrated at
its center
The Cavendish Experiment –
“Weighing the Earth”
“Weighing the Earth” and consequences
• Cavendish didn’t actually weigh the Earth, he
accurately measured the value for G
• Prior to his experiment, the quantities g and RE
were known from direct measurement
Calculate the mass of Earth:
“Weighing the Earth” and consequences
• Calculate the volume of the Earth:
• Calculate the average density of the Earth:
Typical rocks found near the surface of the Earth,
have a density of about 3.00 g/cm3. What can
you conclude?
Acceleration due to gravity
• Acceleration due to gravity (g) varies by location
• g is determined by the distance from the center to
the planet it is near and the mass of the planet
gG
M
r
2
rRh
g as a function of h
Example 12-4
A hypothetical planet has a mass of 2.5 times that of Earth,
but the same surface gravity as Earth. What is this planet’s
radius (in Earth radii)?
Example 12-5
You are explaining to friends why astronauts feel weightless orbiting in
the space shuttle, and they respond that they thought gravity was
just a lot weaker up there. Convince them by calculating how much
weaker gravity is 300 km above Earth’s surface.
Example 12-6
Certain neutron stars (extremely dense stars) are believed to be rotating at
about 1 rev/s. If such a star has a radius of 20 km, what must be its
minimum mass so that material on its surface remains in place during
the rapid rotation?
Some important Astronomy Terms
• Astronomical Unit (AU): is the mean distance from
the Earth to the Sun (1 AU= 1.5×1011 m)
• Periapsis: The point of closest approach of an object
to the body being orbited
– perihelion: closest distance from the Sun
– perigee: closest distance from the Earth
• Apoapsis: The point of furthest excision of an object
from the body being orbited
– aphelion: furthest distance from the Sun
– apogee: furthest distance from the Earth
Kepler’s 1st Law of Orbit – Law of Elliptical Orbits
Planets follow elliptical orbits, with the Sun at one
focus of the ellipse
• The semi-major axis (a) is also equal to the average
distance from a focus to points on the ellipse
Orbital Distance at Apoapsis and Periapsis
• The eccentricity (e) is defined so that ea is the
distance from the center of the ellipse to the
either focus
• In circular orbits the two foci merge to one
focus and e = 0
R a  a 1  e 
R p  a 1  e 
Orbital Shape depends on eccentricity!
• All orbits take the shape of a conic
section, and the shape depends on
the eccentricity
• Closed orbits includes circular orbits
(e = 0) and elliptical orbits (0 < e < 1)
• Open trajectories include parabolas
and hyperbolas where an object
approach the body and will never
return (an escape trajectory)
– Parabolic trajectory (e = 1), object will
reach infinity with zero KE (E = 0)
– Hyperbolic trajectory (e > 1) object will
reach infinity with excess KE (E > 0)
Kepler’s 2nd Law of Orbit – Law of Equal Areas
As a planet moves in its
orbit, it sweeps out an
equal amount of area in
an equal amount of time
• This means that they
move faster when it is
closer to the Sun
• This is a consequence of
the conservation of
angular momentum
v 1 r1  v 2 r2
Example 12-7
The Earth’s orbit around the Sun is in the shape of an ellipse. Its aphelion is
152,000,000 km. Its perihelion, is 147,000,000 km. The Earth’s speed is
30,300 at perihelion. What is the Earth’s orbital speed at aphelion?
Kepler’s 3rd Law of Orbit – Law of Harmonics
The period, T, of a planet is
proportional to its mean
distance from the Sun
raised to the 3/2 power
T  constant
a
or
2 

T
4
 

3
a
GM 
2
3/2
Derive Kepler’s 3rd Law
Example 12-8
Comet Halley orbits the Sun with a with a period of 76 years, and in
1986, had a distance of closest approach to the Sun, its perihelion
distance RP, of 8.9×1010 m. (a) What is Halley’s average distance from
the Sun? (b) What is the eccentricity of comet Halley? (c) What is
the comet’s farthest distance from the Sun, its aphelion distance?
Example 12-9
We SEE nothing at the exact center of our galaxy, the Milky Way. However, we
know that something has to be there, because all of the stars in our galaxy
orbit this point. We can observe a star, called S2 as it moves around this
mysterious object called Sagittarius A*. S2 orbits Sagittarius A* with a period
of 15.2 years and with a semimajor axis of 5.5 light days (= 1.42×1014 m).
What is the mass of Sagittarius A*? What is Sagittarius A*?
Kepler’s 3rd Law (alternate form)
• Because the ratio T2/a3 is a constant for any objects
in orbit around the same mass, we can say
T1
2
3
1
a

T
2
2
3
2
a
 T   a 
1
1
    
T2  a 2 
2
or
3
• When comparing to Earth, remember a = 1 AU and
T = 1 yr
Example 12-10
(a) Uranus has a semi-major axis of 19.23 AU. Calculate Uranus’ orbital
period?
(b) It takes Mercury only 88 days to complete one orbit around the Sun.
Calculate Mercury’s mean distance from the Sun.
Satellite Orbits
• A satellite is said to be in geosynchronous orbit if
it has a period of 1 day
– Such satellites are used in communications and for
weather forecasting
• For a satellite in a circular orbit:
v 
GM
r
Example 12-11
(a) Calculate the orbital speed of a satellite in circular orbit at an altitude
of 1,000 km above the surface of Earth.
(b) Calculate this satellite’s orbital period.
Gravitational Potential Energy at any r
U  G
m 1m 2
r
• Note that U = 0 at infinity, this is the common
convention for problems of an astronomical scale
– Doesn’t have to be, because only differences in U matter
U = mgh when h << R
Energy Conservation
• The mechanical energy E of an object of mass m,
moving at speed v, at a distance from a planet of
mass M is given by
E K U 
1
2
mv  G
2
mM
r
• For a satellite in Kepler orbit, the total energy can
be shown to be:

E  G
mM
2a
Orbital Speed
2 1 
v  GM   
 r a 
2
Example 12-12
An asteroid, headed directly toward Earth, has a speed of 12 km/s relative
to Earth when the asteroid is 10 Earth radii from Earth’s center.
Neglecting the effects of Earth’s atmosphere on the asteroid, find the
asteroid’s speed when it reaches Earth’s surface.
Example 12-13
Venus orbits the Sun with a semi-major axis of 1.082×1011 m and an
eccentricity of 0.068. Calculate Venus’ orbital speed at periheion and
aphelion.
Orbital Maneuvers
• The radius at which a satellite follows a circular
orbital path is directly related to its speed
• Moving to lower orbits requires you to use your
decelerating rockets
– Slows satellite causing it to orbit in an elliptical orbit, a
second rocket burn is required to turn it into a circular
orbit
• Moving to higher orbits requires you to use your
accelerating rockets
The Hohmann Transfer to a lower orbit
The Hohmann Transfer to a higher orbit
Escape Speed
• Escape speed is the minimum speed required to
leave a planet and to never return
– To do this, K = 0 when r = ∞ (where U = 0)
ve 
2GM
R
Example 12-14
Suppose that you fired a cannonball straight upward with an initial
speed equal to one-half the escape speed. How far from the center of
the Earth does this rocket travel before momentarily coming to rest?
(Ignore air resistance in the Earth’s atmosphere)
Example 12-15
What multiple of the energy needed to escape from Earth gives the
energy to escape from (a) the Moon, and (b) Jupiter?
(the moon’s radius is 0.273 Earths and the moon’s mass is 0.0123 Earths,
while Jupiter’s radius is 11.21 Earths and its mass is 317.8 Earths)
Kepler Orbits
Orbital Shape Eccentricity (e)
Circle
e=0
Semi-Major
Axis (a)
a>0
Total Energy
(E)
E<0
Ellipse
0<e<1
a>0
E<0
Parabola
e=1
none
E=0
Hyperbola
e>1
a<0
E>0
Black Holes & the Schwarzchild Radius
• Looking at the escape speed equation: v e  2GM
R
you can seed that escape speed increases
with increasing mass and decreasing radius
• If an object were to be compressed
to a size

small enough that the escape speed necessary
were the speed of light, c then we have a
black hole
RS 
2GM
c
2
Example 12-16
Mr. Bailey has a mass of about 110 kg. How small would you
have to compress him to turn him into a black hole?
Evidence for Black Holes
• Gravitational Lensing
• Motion of stars near it