The Force of

Gravity exists

between any two

masses! Always

attractive… do

you feel the

attraction?

Slide 6-35

Summary

Newton’s law of gravity describes

the gravitational force between

A. the earth and the moon.

B. a person and the earth.

C. the earth and the sun.

D. the sun and the planets.

E. all of the above.

AnswerNewton’s law of gravity describes

the gravitational force between

A. the earth and the moon.

B. a person and the earth.

C. the earth and the sun.

D. the sun and the planets.

E. all of the above.

Checking Understanding:

Gravity on Other Worlds

A 60 kg person stands on each of the

following planets. Rank order her weight on

the three bodies, from highest to lowest.

A. A > B > C

B. B > A > C

C.B > C > A

D.C > B > A

E. C > A > B

Answer: Fg = Gmplanetmperson/r2

A 60 kg person stands on each of the following planets. Rank

order her weight on the three bodies, from highest to lowest.

A.A > B > CB. B > A > C

C. B > C > A

D. C > B > A

E. C > A > B

What is the magnitude and

direction of the force of

gravity on a 60 kg person?

mEarth =5.98 X 1024 kgrEarth =6.37 X 106 m

(Astronomical data in table in back of book.)

Example Problem

Fg = GmEarthmperson/r2

Did you get approximately…

590 N?

And this is an attractive force… Person pulls on Earth with 590 N and Earth pulls on person with 590 N.

A typical bowling ball is spherical,

weighs 16 pounds, and has a diameter

of 8.5 in.(2.21 lbs = 1 kg, 2.54 cm = 1 in)

Suppose two bowling balls are right

next to each other in the rack. What is

the gravitational force between the

two—magnitude and direction?

Example Problem

Fg = Gm1m2/r2

Did you get approximately…

7.5 X 10-8 N?

And this is an attractive force.

Satellites experiencing

Uniform Circular Motion• Circular path is caused by…

• Centripetal Force

• In the case of a satellite orbiting a planet (example: moon around earth), the Fc is caused by the gravitational force

• F = Fc = Fg

• Remember: it is the net force, F, that causes an object to move in UCM

…the satellite’s high speed

The speed of the satellite needs to be very specific because…

too slow… crash into earth (or the object it’s circling).

too fast… leave orbit.

Fig 5.12

What keeps a satellite up?

What is speed, v, for an object

circling earth (or planet in general)? F = Fc = Fg

Fc =mac = mv2/r And since the gravitational force is the only

force causing UCM:

Fg = Gmem/r2

me = mass of earth (or any planet or object that m orbits)

r = distance from center of planet to satellite m = mass of satellite (object experiencing

UCM) G = universal gravitational constant = G= 6.67 X 10-11 Nm2/kg2

Putting it all together… Fc =mac = mv2/r m = mass of satellite, v = speed of satellite, r

= radius of satellite’s path And since the gravitational force is the only

force causing UCM: Fg = Gmem/r2

Fc = Fg = Gmem/r2

mv2/r = Gmem/r2

v2/r = Gme/r2

v2= Gme/r

v = (Gme/r)1/2

Mass of satellite cancels speed of a satellite in orbit does not depend on its mass!

Just like the mass of an object does not affect acceleration in free-fall, neither does mass affect a free-falling satellite in orbit!

In general, speed of a satellite orbiting a

planet of mass mp is given by…

v = (Gmp/r)1/2

This is the only speed the satellite can move at to stay at radius r.

To find or use period, T, in solving:

Recall, v=2∏r/T and v = (Gmp/r)1/2

2∏r/T = (Gmp/r)1/2

[2∏r/T]2 = [(Gmp/r)1/2]2

4∏2r2/T2 = Gmp/r

T2 =4∏2r3/(Gmp)

T = [4∏2r3/(Gmp)]1/2

T = 2∏[r3/(Gmp)]1/2

Period for a satellite orbiting planet of mass mp a distance r from the satellite to the center of planet.

Weightlessness If you sit in your chair, you feel the

normal force (a contact force). But if you are jumping on a trampoline, even while moving through the air, you do not feel the earth pulling upon you with a force of gravity (an action-at-a-distance force). The force of gravity can never be felt. Yet those forces which result from contact can be felt. And in the case of sitting in your chair, you can feel the chair force; and it is this force which provides you with a sensation of weight. Since the upward normal force would equal the downward force of gravity when at rest, the strength of this normal force gives one a measure of the amount of gravitational pull.

If there were no upward normal force acting upon your body, you would not have any sensation of your weight. Without the contact force (the normal force), there is no means of feeling the non-contact force (the force of gravity).

Weightlessness is simply a sensation experienced by an individual when there are no external objects touching one's body and exerting a push or pull upon it. Weightless sensations exist when all contact forces are removed. These sensations are common to any situation in which you are momentarily (or perpetually) in a state of free fall. When in free fall, the only force acting upon your body is the force of gravity - a non-contact force. Since the force of gravity cannot be felt without any other opposing forces, you would have no sensation of it. You would feel weightless when in a state of free fall.

Creating Artificial Gravity The surface of the

rotating space station pushes on an object with which it is in contact and thereby provides the centripetal force that keeps the object moving on a circular path

Fig 5.19

At what speed must the surface of the space station (r=1700m) move so that the astronaut at point P experiences a push on his feet that equals his earth weight?

The floor of the rotating space station exerts a normal force on the feet of the astronaut.

This is the centripetal force that keeps the astronaut moving in a circular path.

If the magnitude of the normal force must equal the astronaut’s earth weight, then…

Fc = mg = mv2/r

Solve for v:

v = (rg)1/2 = [(1700m)(9.8m/s2)]1/2

v=130 m/s

How large (radius =?) would a space station

need to be to simulate 1g? How fast would

you be traveling on this space station? People tend to get sick if rotation is greater than 1rpm,

so let f=1rpm; T=60s/rev

1g=9.80m/s2

Recall: ac = v2/r and v = 2∏r/T

r= acT2/(4∏2)

r=(9.80 m/s2)(60 s)2/(4∏2)

r=893.65m

v = 2∏r/T = 2∏(893.65m)/(60s)

v=93.58 m/s

v=209 mi/hr

Example Problem: Orbital Motion

Phobos is one of two small

moons that orbit Mars. Phobos

is a very small moon, and has

correspondingly small

gravity—it varies, but a typical

value is about 6 mm/s2.

Phobos isn’t quite round, but it

has an average radius of

about 11 km. What would be

the orbital speed around

Phobos, assuming it was

round with gravity and radius

as noted?

Slide 6-34

Answer

A spacecraft is orbiting the moon in an orbit

very close to the surface—possible because

of the moon’s lack of atmosphere. What is

the craft’s speed? The period of its orbit?

Phobos is the closer of Mars’ two small

moons, orbiting at 9400 km from the center of

Mars, a planet of mass

6.4 1023 kg. What is Phobos’ orbital period?

How does this compare to the length of the

Martian day, which is just shy of 25 hours?

Example Problems: Gravity and Orbits

Answer