9.4.1State that gravitation provides the centripetal force for circular orbital motion. 9.4.2Derive...

9.4.1 State that gravitation provides the centripetal force for circular orbital motion.

9.4.2 Derive Kepler’s third law. The third law states that the period squared of a planet in orbit is proportional to the cube of the radius of its orbit.

Topic 9: Motion in fields9.4 Orbital motion

State that gravitation provides the centripetal force for circular orbital motion.

Consider a baseball in circular orbit about Earth.

Clearly the only force that is causing the ball to move in a circle is the gravitational force.

Thus the gravitational force is the centripetal force for circular orbital motion.EXAMPLE: A centripetal force causes a centripetal acceleration ac. What are the two forms for ac?

SOLUTION: Recall from Topic 2 that ac = v2/r.

Then from the relationship v = 2r/T we see that ac = v2/r = (2r/T)2/r = 42r2/(T2r) = 42r/T2.


EXAMPLE: Suppose a 0.500-kg baseball is placed in a circular orbit around the earth at slightly higher that the tallest point, Mount Everest (8850 m). Given that the earth has a radius of RE = 6400000 m, find the speed of the ball.

SOLUTION: The ball is traveling in a circle of radius r = 6408850 m.

Fc is caused by the weight of the ball so that

Fc = mg = (0.5)(9.8) = 4.9 n.

Since Fc = mv2/r we have

4.9 = (0.5)v2/6408850

v = 7925 m s-1!



FYIWe assumed that g = 10 ms-2 at the top of Everest.

http://de.wikipedia.org/w/index.php?title=Datei:Everest_North_Face_toward_Base_Camp_Tibet_Luca_Galuzzi_2006.jpg&filetimestamp=20070320004704

PRACTICE: Find the period T of one complete orbit of the ball.SOLUTION: r = 6408850 m.Fc = 4.9 n.Fc = mac = 0.5ac so that ac = 9.8.But ac = 42r/T2 so that T2

= 42r/ac

T2 = 42(6408850)/9.8

T = 5081 s = 84.7 min = 1.4 h.



ac = v2/r centripetal accelerationac = 42r/T2


Derive Kepler’s third law. The third law states that the period squared of a planet in orbit is proportional to the cube of the radius of its orbit.


EXAMPLE: Show that for an object in a circular orbit about a body of mass M that T2 = (42/GM)r3.

SOLUTION:

In circular orbit Fc = mac and Fc = GMm/r2.

But ac = 42r/T2. Then

mac = GMm/r2

42r/T2 = GM/r2

42r3 = GMT2

T2 = [42/(GM)]r3

FYIThe IBO expects you to be able to derive this relationship.

PRACTICE: Using Kepler’s third law find the period T of one complete orbit of the baseball from the previous example.SOLUTION: r = 6408850 m.G = 6.67×10−11 N m2

kg−2.M = 5.98×1024 kg.Then T2 = (42/GM)r3 so thatT2

= [42/(6.67×10−11×5.98×1024)](6408850)3

T = 5104 s = 85.0 min = 1.4 h.



T2 = [42/(GM)]r3 Kepler’s third law

FYINote the slight discrepancy in the period (it was 5081 s before). How do you account for it?


9.4.3 Derive expressions for the kinetic energy, potential energy and total energy of an orbiting satellite.

9.4.4 Sketch graphs showing the variation with orbital radius of the kinetic energy, gravitational potential energy and total energy of a satellite.

9.4.5 Discuss the concept of weightlessness in orbital motion, in freefall and in deep space.

9.4.6 Solve problems involving orbital motion.


Derive expressions for the kinetic energy, potential energy and total energy of an orbiting satellite.

An orbiting satellite has both kinetic energy and potential energy.

As we learned in Topic 9.2, the gravitational potential energy of an object of mass m in the gravitational field of Earth is EP = -GMm/r.

As we learned in Topic 2.3, the kinetic energy of an object of mass m moving at speed v is EK = (1/2)mv2.

Thus the total mechanical energy of an orbiting satellite of mass m is

Where M is the mass of the earth.


E = EK + EP total energy of an

orbiting satelliteE = (1/2)mv2 - GMm/r



EXAMPLE: Show that the kinetic energy of an orbiting satellite at a distance r from the center of Earth is EK = GMm/(2r).

SOLUTION:

In circular orbit Fc = mac and Fc = GMm/r2.

But ac = v2/r. Then

mac = GMm/r2

mv2/r = GMm/r2

mv2 = GMm/r

(1/2)mv2 = GMm/(2r)kinetic energy of an

orbiting satelliteEK = (1/2)mv2 = GMm/(2r)



EXAMPLE: Show that the total energy of an orbiting satellite at a distance r from the center of Earth is E = -GMm/(2r).

SOLUTION: From E = EK + EP and the expressions for

EK and EP we have E = EK

+ EP

E = GMm/(2r) - GMm/r

E = GMm/(2r) - 2GMm/(2r)

E = -GMm/(2r)

FYIIBO expects you to derive these relationships.

total energy of an orbiting satellite

E = -GMm/(2r)EK = GMm/(2r) EP = -GMm/r

Sketch graphs showing the variation with orbital radius of the kinetic energy, gravitational potential energy and total energy of a satellite.




EXAMPLE: Graph the kinetic energy vs. the radius of orbit for a satellite of mass m about a planet of mass M and radius R.

SOLUTION: Use EK = GMm/(2r). Note that EK

decreases with radius. It has a maximum value of EK = GMm/(2R).

EK

rR 2R 3R 4R 5R

GMm2R





EXAMPLE: Graph the potential energy vs. the radius of orbit for a satellite of mass m about a planet of mass M and radius R.

SOLUTION: Use EP = -GMm/r. Note that EP increases with radius. It becomes less negative.

EP

rR 2R 3R 4R 5R

GMmR

-





EXAMPLE: Graph the total energy E vs. the radius of orbit and include both EK and EP.

SOLUTION:

GMmR

-

ErR 2R 3R 4R 5R

GMm2R

-

GMm2R

+

EK

EP

PRACTICE: If the elevator were to accelerate upward at 2 ms-2, what would Dobson observe the dropped ball’s acceleration to be?SOLUTION: Since the elevator is accelerating upward at 2 ms-2 to meet the ball which is accelerating downward at 10 ms-2, Dobson would observe an acceleration if 12 ms-2.If the elevator were accelerating downward at 2, he would observe an acceleration of 8 ms-2.

Discuss the concept of weightlessness in orbital motion, in freefall and in deep space.

Consider Dobson inside an elevator which is not moving…

If he drops a ball, it will accelerate downward at 10 ms-2 as expected.


PRACTICE: If the elevator were to accelerate downward at 10 ms-2, what would Dobson observe the dropped ball’s acceleration to be?SOLUTION: He would observe the acceleration of the ball to be zero!He would think that the ball was “weightless!”

FYIThe ball is NOT weightless, obviously. It is merely accelerating at the same rate as Dobson!How could you get Dobson to accelerate downward at 10 ms-2? Cut the cable!



The “Vomit Comet”

PRACTICE: We have all seen astronauts experiencing “weightlessness.” Explain why it only appears that they are weightless.SOLUTION: The astronaut, the spacecraft, and the tomatoes, are all accelerating at ac = g.They all fall together and appear to be weightless.



International Space Station


In deep space, the r in F = GMm/r2 is so large for every m that F, the force of gravity, is for all intents and purposes, zero.


Only in deep space – which is defined to be far, far away from all masses – will a mass be truly weightless.

Solve problems involving orbital motion.


KE is POSITIVE and decreasing.

GPE is NEGATIVE and increasing (becoming less negative).



T2 = (42/GM)r3 Kepler’s third law

T2 = [42/(GM)]r3 r3 = [GM/(42)]T2.



g = GM/r2 (from Topic 9.2).EK = GMm/(2r) = (1/2)mv2.

GM/r = v2.GM/r2 = v2/r

g = v2/r



R

x

g = GM/x2 (from Topic 9.2).

ac = GM/x2 (since ac = g in circular orbits).

v2/x = GM/x2 (since ac = v2/r).

v2 = GM/x so that v = GM/x.



R

x

But EK = (1/2)mv2.

Thus EK = (1/2)mv2 = (1/2)m(GM/x) = GMm/(2x).

Use EP = mV and V = -GM/x. Then EP = m(-GM/x) = -GMm/x.

From (a) v2 = GM/x.



R

x

E = EK + EP

E = GMm/(2x) + -GMm/x [ from (b)(i) ]

E = 1GMm/(2x) + -2GMm/(2x)

E = -GMm/(2x)



R

x

The satellite will begin to lose some of its mechanical energy in the form of heat.



R

x

Refer to E = -GMm/(2x) [ from (b)(ii) ]:If E gets smaller x must also get smaller (since E is negative).If r gets smaller the atmosphere must get thicker and more resistive.Clearly the orbit will continue to decay (shrink).




T2 = (42/GM)r3

T = [(42/GM)r3]1/2

T = (42/GM)1/2r3/2

T r3/2



M2M1

R1

R2

P

It is the gravitational force.



M2M1

R1

R2

P

Note that FG = GM1M2/(R1+R2)2.M1 experiences Fc = M1v1

2/R1.Since v1 = 2R1/T, v1

2 = 42R12/T2.

Then Fc = FG M1v12/R1 = GM1M2/(R1+R2)2.

M1(42R12/T2)/R1 = GM1M2/(R1+R2)2

42R1(R1+R2)2 = GM2T2

T2 = R1(R1+R2)242

GM2



M2M1

R1

R2

P

From (b) T2 = (42/GM2)R1(R1+R2)2.From symmetry T2 = (42/GM1)R2(R1+R2)2.

(42/GM2)R1(R1+R2)2 = (42/GM1)R2(R1+R2)2

(1/M2)R1 = (1/M1)R2

M1/M2 = R2/R1Since R2 > R1, M1 > M2.





If r decreases EK get bigger.If r decreases EP get more negative (smaller).




TX2 = (42/GM)rX

3.TY

2 = (42/GM)rY3.

TX = 8TY TX2 = 64TY

2.

TX2/TY

2 = (42/GM)rX3/[(42/GM)rY

3]

64TY2/TY

2 = rX3/rY

3

64 = (rX/rY)3

rX/rY = 641/3 = 4

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9.4.1State that gravitation provides the centripetal force for circular orbital motion. 9.4.2Derive...

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