1. The story of the apple
When Newton observed the apple fall, he wondered if the force that caused the apple to fall to the ground was the same force that kept celestial bodies in orbit

2. Universal Gravitation
• Newton’s Law of Universal Gravitation states that all objects with mass exert attractive forces on all other objects with mass.
• The force is proportional to the product of the masses and inversely proportional to the square of the separation distance
Where:
Fg = Force due to
gravity (N)
m1 & m2 = mass of
each body (kg)
r = separation distance (m)
G = 6.67 x Nm2/kg2

3. HENRY CAVENDISH
Thin flat wire
In 1797 Henry Cavendish devised a very sensitive torsion balance to study gravitational attraction. He was the first to observe gravity in the lab. He actually measured the force which existed between two known masses, and used this data to calculate a proportionality constant.
Twist in wire measures force F
G = 6.67 x Nm2/kg2 F

4. Understanding Universal Gravitation
All objects experience a force of gravity that points toward the center of the earth, even when they are not in contact.
Does the moon
experience a force of gravity as well?
of course

5. Understanding Universal Gravitation
The force of gravity on an
object is proportional to
the mass of the object
The force of gravity is
proportional to the mass of
the object exerting the
force.
The force of gravity is
inversely proportional to
the square of the separation distance.
G = 6.67 x Nm2/kg2

6. Determine the force of attraction between the earth and the moon
What do we need to know?
m1 = mass of the moon = 7.36 × 1022 kg m2 = mass of the earth = 5.97 × 1024 and r = distance between the earth and the moon = 384,402 km
Fg = 1.99 x 1026 N

7. m = 9.0 x 10-31 kg The gravitational force between two electrons
1.00 m apart is 5.42 x N. What is the mass
of an electron?
F= 5.42 x N
r = 1.00 m m1 = m2 = ?
m1 = m2 (m)(m) = m2
m = 9.0 x kg

8. Satellites • To stay moving in a circle, the object must experience
a centripetal force.
• Gravity provides the
centripetal force.

9. The Andromeda galaxy, known as M31, is 2. 1 x 106 light years away
The Andromeda galaxy, known as M31, is 2.1 x 106 light years away. Measurements show that a star out at its extremities 7.5 x 1020 AU (1.125 x 1032 m) from the center orbits the center at 200,000 m/s. Approximate the mass of M31. All the mass can be taken to act at the center of gravity.
r = x 1032 m
v = 200,000 m/s m = ?
m = 6.7 x 1052 kg

10. KEPLER’S THREE LAWS OF PLANETARY MOTION
First Law
The planets orbit the sun in elliptical paths with the sun at one of the foci

11. Second Law
A planet sweeps out equal areas in equal time intervals as it proceeds around the sun
Naturally the real implication of this law is what:
Planets move faster when closer to the sun and slower when farther away

12. Third Law
The ratio of the cube of the orbital radius to the square of the
orbital period is constant for all planets
R R23
_____________ ____________ =
T T22
or R3 Gm
____________ __________________ =
T2 42
Where m is the mass of the central body.

13. What is the period of Venus’ orbit?
Looking up the data for Venus on the chart, we find
that Venus is 0.72 AU from the sun. This is its orbital
radius R2.
Since Venus and the earth both orbit the same central
body (the sun), we can apply Kepler’s Third Law.
Given: R1 = 1 AU (Earth)
T1 = 1 year (Earth)
R2 = 0.72 AU (Venus)
T2 = ? (Venus)
T2 = 0.6 years = 223 days

14. Given: m1 = 150 kg m2 = 5.98 x 1024 kg (Earth)
A 150 kg satellite is launched into orbit 12,800 km above the earth’s surface. What is the weight of the satellite on earth? What is the weight of the satellite while in orbit? What is the speed of the satellite while in orbit? What is its orbital period?
Given: m1 = 150 kg m2 = 5.98 x 1024 kg (Earth)
radius of Earth = km
altitude of satellite above Earth = 12,800 km
radius of satellite's orbit = = 19,170 km = 19,170,000 m
Fg = mg
Fg = (150)(9.8) = 1470 N on Earth*
* Of course you could have used Newton’s Law of Universal Gravitation
with the radius of the Earth as r and arrived at the same answer.

15. Fg = 163 N while in orbit
v = 4,564 m/s

16. = v = 4,564 m/s r = 19,170,000 m T = 26,400 seconds = 7.3 hours OR
R Gm
___________ ____________ =
(19,170,000) (6.67 x 10-11)(5.98 x 1024)
___________________________ __________________________________________ =
T 2
T 2