SP221/2141&4341
Chapter 11 Homework
Due Friday December 2, 2016
Question 1: A satellite of mass m=1500 kg orbits the Earth in a circular orbit at an altitude of
4800 km above the Earth’s surface. How much energy is required to lift it into a circular orbit at an
altitude of 9600 km above the Earth’s surface? (Note: the speed of the satellite will need to change.)
[5 points]
Question 2: Suppose that a tunnel is drilled to the center of the Earth. Assume that this tunnel
somehow does not fill up with water, or molten iron from the Earth’s core or anything else. Also
ignore air resistance, the Earth’s rotation, and friction with the tunnel walls. You drop an object
into the tunnel. (a) How fast is it moving when it reaches the center of the Earth? (b) What is
the gravitational potential energy of a 1 kg object located at the center of the Earth, defining the
potential energy to be zero if the object were infinitely far from the Earth. (c) Draw an accurate
graph showing the magnitude of the gravitational potential energy of a 1 kg object as a function of
r, its distance from the center of the Earth. Make the plot run from r=0 up to at least r = 3RE .
[15 points]
Question 3: An asteroid of mass m=2.5×1013 kg is moving as shown below. It has an initial
speed of v0 =3700 m/s when it is very far away (effectively infinitely far away) from the Earth.
The asteroid just barely strikes the Earth with a glancing (tangential) blow. The distance shown
as h in the figure is known as the asteroid’s impact parameter. (This is the minimum for the
component of the distance between the Earth and the asteroid perpendicular to asteroid’s velocity
such that it will strike the Earth.) What is this distance? [Hint: Start by determining how fast
the asteroid will be moving when it reaches the Earth’s surface. The only force acting on the
asteroid is gravitational attraction to the Earth, so there will be no torque acting on the asteroid
about the center of the Earth. You can therefore use conservation of angular momentum.] [10 points]
v0 =3700 m/s
h
Earth
SP221/2141&4341
Chapter 11 Homework
Due Friday December 2, 2016
Question 4: At present, the Earth rotates on its axis with a period (relative to the fixed stars) of
one sidereal day: TE =8.61×104 s. The moon orbits the Earth with a period of one lunar month:
Tm =27.3 days=2.36×106 s. However, tidal forces between the Earth and moon will affect these.
These forces have already caused the moon to become tidally locked with the Earth - the moon’s
period of rotation about its own axis is equal to its period of rotation about the Earth so that the
same side of the moon is always facing the Earth. Over a very long period of time the same thing
will happen to the Earth: tidal friction will dissipate energy, the period of rotation of both the
Earth and moon will get longer, and the moon will move farther away from the Earth. Eventually
the length of a “day” will equal the length of a “lunar month” so that the same side of the Earth
will always face the moon. How much energy, in joules, will be dissipated by tidal forces in this
process? The mass of the Earth (ME ), the radius of the Earth (RE ), the mass of the moon (m),
and the current Earth-moon distance are given at the start of your textbook. Treat the Earth
as a sphere of uniform density. Treat the moon as a point particle, and ignore its spin angular
momentum about its axis. [20 points]
(a) The distance between the Earth and moon will change, so we will want to eliminate it from our
equations. So as a first step, solve for the radius of the moon’s orbit in terms of G, ME , and Tm .
(b) Write down the total initial angular momentum of the Earth-moon system. Write it in terms
of the given quantities: G, ME , RE , TE , m, and Tm . To simplify the equation, you can also use
quantities derived from these: Ie (the moment of inertia of the Earth about its axis), and what I’ll
E 1/3
call α = ( GM
)
4π 2
(c) Write the the total final angular momentum of the Earth-moon system after they are in
equilibrium. Write your answer in terms of Ie , α, m, and Tf where Tf is the final rotational period
of both the Earth and moon.
(d) Use conservation of angular momentum to solve for Tf . Then convert Tf into (current) days.
The exact equation will not be easy to solve in closed form. If you want, you can use your calculator
to solve it numerically. There will likely be two answers: we want the answer that is longer than a
current day (the shorter answer does conserve angular momentum, but would require a significant
increase in mechanical energy). Alternatively, we can assume that in the final configuration when
the Earth and moon have the same period the angular momentum due to the Earth will be much
smaller than that due to the moon. (This should be obvious from comparing their moments of
inertia.) So you can drop the term corresponding to the Earth’s angular momentum to make the
equation solvable for Tf .
(e) Once you have Tf , use this to solve for the final (center to center) distance between the Earth
and the moon. Give this answer both in meters and as a multiple of the current distance.
(f) Determine the change in mechanical energy of the Earth-moon system. Consider the change
in gravitational potential energy, the change in the Earth’s kinetic energy, and the change in the
moon’s kinetic energy.