The Force of Gravity exists between any two masses! Always attractive do you feel the attraction? Slide 6-35

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.

Answer 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.

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: F g = Gm planet m person /r 2 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

Example Problem What is the magnitude and direction of the force of gravity on a 60 kg person? m Earth =5.98 X 10 24 kg r Earth =6.37 X 10 6 m (Astronomical data in table in back of book.)

F g = Gm Earth m person /r 2 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.

Example Problem 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?

F g = Gm 1 m 2 /r 2 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 F c is caused by the gravitational force F = F c = F g Remember: it is the net force, F, that causes an object to move in UCM

What keeps a satellite up? 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 is speed, v, for an object circling earth (or planet in general)? F = F c = F g F c =ma c = mv 2 /r And since the gravitational force is the only force causing UCM: F g = Gm e m/r 2 m e = 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 Nm 2 /kg 2

Putting it all together F c =ma c = mv 2 /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: F g = Gm e m/r 2 F c = F g = Gm e m/r 2 mv 2 /r = Gm e m/r 2 v 2 /r = Gm e /r 2 v 2 = Gm e /r v = (Gm e /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 m p is given by v = (Gm p /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 = (Gm p /r) 1/2 2 r/t = (Gm p /r) 1/2 [2 r/t] 2 = [(Gm p /r) 1/2 ] 2 4 2 r 2 /T 2 = Gm p /r T 2 =4 2 r 3 /(Gm p ) T = [4 2 r 3 /(Gm p )] 1/2 T = 2 [r 3 /(Gm p )] 1/2 Period for a satellite orbiting planet of mass m p 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-adistance 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 noncontact 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 F c = mg = mv 2 /r Solve for v: v = (rg) 1/2 = [(1700m)(9.8m/s 2 )] 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/s 2 Recall: a c = v 2 /r and v = 2 r/t r= a c T 2 /(4 2 ) r=(9.80 m/s 2 )(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/s 2. 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

Example Problems: Gravity and Orbits 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 10 23 kg. What is Phobos orbital period? How does this compare to the length of the Martian day, which is just shy of 25 hours?

Answer