Physics 18 Spring 2011

Homework 9 - Solutions

Wednesday March 16, 2011

Make sure your name is on your homework, and please box your final answer. Becausewe will be giving partial credit, be sure to attempt all the problems, even if you don’t finishthem. The homework is due at the beginning of class on Monday, March 30th. Becausethe solutions will be posted immediately after class, no late homeworks can be accepted! Youare welcome to ask questions during the discussion session or during office hours.

1. Why is the gravitational potential energy of two masses negative? Note that saying“because that’s what the equation gives” is not an explanation!

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Solution

The gravitational potential energy represents the energy of the system, which you canthink of as the amount of work that it took to put the system together. Because themasses attract each other gravitationally, then they naturally fall together. This meansthat the work was done for us, and so the system has negative energy. This is also theamount of energy that we’d need to add to the system to break it apart. Since we’dneed to do work on the system to break it apart (ending up with zero total energy), wehad to start with a negative energy. The same thing occurs with oppositely-chargedparticles like protons and electrons, which also have negative electrostatic potentialenergy.

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2. Some people think that the shuttle astronauts are “weightless” because they are “be-yond the pull of Earth’s gravity.” In fact, this is completely untrue.

(a) What is the magnitude of the gravitational field in the vicinity of a shuttle orbit?A shuttle orbit is about 400 km above the ground.

(b) Given the answer in Part (a), explain why shuttle astronauts suffer from ad-verse biological effects such as muscle atrophy even though the are not actually“weightless.”

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Solution

(a) The shuttle orbits at a distance of r = RE +h from the center of the Earth, whereRE is the radius of the Earth, and h is the height above the surface. If the heightis h = 400 km, then r = 6400 + 400 = 6800 km, or 6.8 × 106 meters. At thispoint, the gravitational field has a magnitude

g =GME

r2=

6.67× 10−11 × 5.98× 1024

(6.8× 106)2= 8.6 m/s2,

which is still almost 90% of the acceleration at the surface of the Earth.

(b) Remember that the weight that we feel is due to the normal force. The astronautsin orbit are in constant free-fall, and don’t feel their weight. So, without thecompensating normal force to fight against the muscles begin to weaken, notneeding to do as much anymore.

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3. Calculate the mass of Earth from the period of the moon, T = 27.3 d; its mean orbitalradius rm = 3.84× 108 m; and the known value of G.

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Solution

The moon is held to the Earth by the gravitational force, FG = GmmoonMEarth

r2 , wherer is the orbital radius. Now, because the orbit is (pretty much) circular, then the netforce on the moon is the centripetal force, Fcent = mmoonv2

r. But, for an object moving

in a circle, then v = 2πr/T , and so

Fcent =mmoon

r

(2πr

T 2

)2

=4π2

T 2mmoonr.

Because the orbit is stable this force is equal to the gravitational force. Setting thetwo forces equal gives

4π2

T 2mmoonr = G

mmoonMEarth

r2.

Solving for the mass of the Earth gives

MEarth =4π2

T 2Gr3.

The period is T = 27.3× 3600× 24 = 2.36× 106 seconds, and so

MEarth =4π2

T 2Gr3 =

4π2

(2.36× 106)2(6.672× 10−11)(3.84× 108)3 = 6× 1024 kg.

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4. (a) If we take the potential energy of a 100 kg object and Earth to be zero when thetwo are separated by an infinite distance, what is the potential energy when theobject is at the surface of Earth?

(b) Find the potential energy of the same object at a height above Earth’s surfaceequal to Earth’s radius.

(c) Find the escape speed for a body projected from this height.

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Solution

(a) The gravitational potential energy of two objects of masses m and M , separatedby a distance r is

PEgrav = −GmMr

,

which sets the potential energy to zero when the objects are infinitely far apart(r → ∞). So, if the object is on the surface of the Earth, then r = RE, andM = ME, and

PEg = −GmME

RE

.

Now, we can rewrite this recalling that for Earth, g = GME

R2E

. Thus,

PEg = −GmME

RE

= −mGME

R2E

RE = −mgRE.

Now, if the radius of the Earth is 6400 km, or 6.4× 106 meters, then

PEg = −mgRE = −(100)(9.8)(6.4× 106) = −6.3× 109 J.

(b) The potential energy at a distance of r = RE +RE = 2RE will just be

PEg = −GmME

2RE

= −1

2mgRE = −3.15× 109 J.

(c) To determine the escape velocity from this distance we just recall the usual methodsetting KE + PE = 0, so 1

2mv2 = 1

2mgRE, giving

vesc =√gRE =

√(9.8)(6.4× 103) = 8 km/s,

which is smaller than the escape velocity from the surface of the earth (vesc ≈ 11km/s), as we should expect.

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5. Black holes are objects whose gravitational field is so strong that not even light canescape. One way of thinking about this is to consider a spherical object whose densityis so large that the escape speed at its surface is greater than the speed of light, c. If astar’s radius is smaller than a value called the Schwarzschild radius RS, then the starwill be a black hole, that is, light originating from its surface cannot escape.

(a) For a nonrotating black hole, the Schwarzschild radius depends only upon the massof the black hole. Show that it is related to that mass M by RS = (2GM) /c2.

(b) Calculate the value of the Schwarzschild radius for a black hole whose mass is tensolar masses.

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Solution

(a) The escape velocity of a spherical mass M with radius R is given by

vesc =

√2GM

r.

If we want light to be unable to escape from the surface, then the escape velocityhas to be greater than, or equal to, the speed of light. Thus, setting the escapevelocity vesc = c and solving for the radius gives

RS =2GM

c2.

(b) The mass of the sun, Msun = 2×1030 kg. So, we can determine the Schwarzschildradius for this black hole,

RS =2GM

c2=

2× 6.67× 10−11 × 10× 2× 1030

(3× 108)2= 29.6 km.

So, the mass of ten suns is squashed into a ball about the size of a large city.

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