Newton’s Law of Universal GravitationEvery object in the universe is attracted to every other object.

F= Gm1m2

r2

G = 6.67 X 10-11 N-m2/kg2

m1 = mass of one object

m2 = mass of second object

r = distance from center of objects

Cavendish proves the law in 1798

Everything in the solar system pulls on everything else.

Sun pulls on Earth All the other planets also pull on the Earth

a. Calculate the force of gravity between two 60.0 kg (132 lbs) people who standing 2.00 m apart. (6.00 X 10-8 N)

b. Calculate the force of gravity between a 60 kg person and the earth? Assume the earth has a mass of 5.98 X 1024 kg and a radius of 6,400,000 m (4,000 miles). (584 N)

A planet has a mass four times that of the earth, but the acceleration of gravity is the same as on the earth’s surface. Relate the planet’s radius to that of the earth (2Re).

A 2000-kg satellite orbits the earth at an altitude of 6380 km (the radius of the earth)above the earth’s surface. What is the force of gravity on the satellite?

F= Gm1m2 = (6.67 X 10-11 N m2/kg2)(2000kg)(5.98 X 1024 kg)

r2 (6,380,00 m + 6,380,00 m)2

F = 4900 N

A simpler solution:

At the earth’s surface:FG = mg

At twice that distance:FG = (1/2)2mg = ¼ mg

FG = (¼ )(2000 kg)(9.80 m/s2)

FG = 4900 N

What is the net force on the moon when it is at a right angle with the sun and the earth?

Relevant Data:MM = 7.35 X 1022 kg

ME = 5.98 X 1024 m

MS = 1.99 X 1030 m

rMS = 1.50 X 1011 m

rME = 3.84 X 108 m

Calculate each force separately:

FME = 1.99 X 1020 N

FMS = 4.34 X 1020 N

FR2 = FME

2 + FMS2

FR = 4.77 X 1020 N

tan = opp = FME

adj FMS

= 24.6o

Earth

Sun

FME

FMS

FR

Consider an object (any object)

Weight = mg (Force due to gravity)mg = GmmE

rE2

g = GmE

rE2

mE =g rE2 = (9.8 m/s2)(6.38 X 106 m)2 = 5.98 X 1024 kg

G 6.67 X 10-11 N-m2/kg2

Calculating “g”g = GmE

rE2

g = (6.67 X 10-11 N-m2/kg2)(5.98 X 1024 kg)(6.38 X 106 m)2

g = 9.80 m/s2

Calculate the value of g at the top of Mt. Everest, 8848 m above the earth’s surface.

g = GmE

r2

g = (6.67 X 10-11 N-m2/kg2)(5.98 X 1024 kg)(6.38 X 106 m + 8848 m)2

g = 9.77 m/s2

g varies with:• Altitude• Location

– Earth is not a perfect sphere– Different mineral deposits can change density– “salt domes” are low density salt regions near

petroleum deposits

Gravitational Potential Energy• mgh is not valid except near the surface of the earth• Zero point is an infinite distance from the earth • Need calculus since the force varies with distance

U = - ∫F dxU = - ∫ Gm1m2 dr

r2

U = -Gm1m2

r

A 1000 kg rocket is launched from the surface of the earth. Calculate the escape velocity. Hint: set the final K and U equal to zero. (11,200 m/s)

A 1000 kg rocket is launched from the surface of the moon. Calculate the escape velocity. The mass of the moon is 7.35 X 1022 kg and the radius is 1.74 X 106 m. (2383 m/s)

A tile of a rocket falls off at an altitude of 1500 km above the surface of the earth. The rocket is travelling upwards at 2000 m/s.

a. Calculate the speed of the tile as it hits the ground. (5276 m/s)

b.In reality. Could the tile actually reach this speed?

Suppose the earth stops orbiting and falls into the sun (see page 395)

a. Use conservation of energy to calculate the speed of the earth as it just crashes into the sun. (6.13 X 105 m/s)

b.Why doesn’t our earth get pulled into the sun?

What speed should a satellite be launched if it needs to have a speed of 500 m/s at 400 km. Remember to include the radius of the earth in your calculations. (2755 m/s)

Why don’t satellites fall back onto the earth?

• Speed• They are “falling”

due to the pull of gravity

• Can feel “weightless” if you are on board (just like in the elevator)

Speed of a SatelliteF= GMm = mv2

r2 r

GM = v2

r

v = \/GM/r

The Starship Enterprise wishes to orbit the earth at a 300 km height.

a. Calculate the total distance of the Enterprise from the center of the earth. (6.68 x 106 m)

b.Calculate the proper orbital speed for the Enterprise. (7730 m/s)

c. Would a heavier starship require a greater orbital speed?

Kepler’s Laws (1571-1630)

1. The orbit of each planet is an ellipse, with the sun at one focus

2. Each planet sweeps out equal areas in equal time

3. T1 2 = r1 3

T2 r2

1. The orbit of each planet is an ellipse, with the sun at one focus

http://csep10.phys.utk.edu/astr161/lect/history/kepler.html

2. Each planet sweeps out equal areas in equal time

• Suppose the travel time in both cases is three days.

• Shaded areas are exactly the same area

Mars has a year that is about 1.88 earth years. What is the distance from Mars to the Sun, using the Earth as a reference (rES = 1.496 X 108 m)

T12 = r1

3

T22 r2

3

TM2 = rM

3

TE2 rE

3

rM3 = TM

2rE3

TE2

rM3 = (1.88y)2(1.496 X 108 m)3

(1 y)2

rM3 = 1.18 X 1025 m3

rM = 2.28 X 108 m

How long is a year on Jupiter if Jupiter is 5.2 times farther from the Sun than the earth?

TJ2 = rJ

3

TE2 rE

3

TJ2 = rJ

3 TE2

rE3

TJ2 = rJ

3 TE2 = (5.2)3(1 y)2

rE3 (1)3

TJ2 = 141 y2

TJ = 11.8 y

How high should a geosynchronous satellite be placed above the earth? Assume the satellite’s period is 1 day, and compare it to the moon, whose period is 27 days.

Ts2 = rs

3

TM2 rM

3

rs3 = rM

3 Ts2

TM2

rs3 = rM

3 Ts2 = rM

3 (1 day)2

TM2 (27 day)2

rS3 = rM

3 Take the 3rd root of

729 both sides

rs = rM

9

The satellite must orbit 1/9 the distance to the moon.

Deriving the Third LawTo derive Kepler’s Law, we will need two

formulas.F= Gm1mJ

r2

F=m1v2

rmJ m1

GMm = mv2

r2 rGM = v2 Substitute v=2r

r TGM = 42r2

r T2

T2 = 42

r3 GM

T2 = 42 We can do this for two

r3 GM different moonsT1

2 = 42 T22 = 42

r13 GM r2

3 GMT1

2 = T22

r13 r2

3

A Useful Form

This form of the equation:

T2 = 42

r3 GM

Is useful for determining the mass of the central planet, using only the period and distance of one of the satellites.

S could be the Sun, Earth, or other body with satellites.

Calculate the mass of the sun, knowing that the earth is 1.496 X 108 m from the sun. (2.0 X 1030 kg)

A student is given the following data and asked to calculate the mass of Saturn. The data describes the orbital periods and radii of several of Saturn’s moons.

Orbital Period, T Orbital Radius, R(seconds) (m)8.14 X 104 1.85 X 108

1.18 X 105 2.38 X 108

1.63 X 105 2.95 X 108

2.37 X 105 3.77 X 108

4p2

Let’s use this equation:

T2 = 42

r3 GmS

And rearrange it:GmS= 142 r3 T2

Once more:1 = GmS

T2 42 r3

Calculate the following values and graph them.

1 G T2 42 r3

0.00E+00

2.00E-11

4.00E-11

6.00E-11

8.00E-11

1.00E-10

1.20E-10

1.40E-10

1.60E-10

0.00E+00 5.00E-38 1.00E-37 1.50E-37 2.00E-37 2.50E-37 3.00E-37

Calculate the slope of the graph

y = m x + b

1 = mS GT2 42 r3

ms = 5.9 X 1026 kg

Orbital EnergiesGMm = mv2

r2 r

GMm = mv2

r 1 mv2 = GMm = K2 2r K = -½ Ug

E = K + Ug

E = - ½ Ug + Ug

E = ½ Ug

A 1500 kg satellite orbits at an altitude of 4,000 km above the earth’s surface. Calculate the work required to move that satellite to an orbit of 12,000 km above the earth’s surface. (1.25 X 1010 J)