Chapter 12: LAW of GRAVITY-Answers & Hints
(Updated 3/26/12)
If you need astronomical data you should check the Appendix in the back of the text. In these problems
you should use the proper value of g=9.8 m/s2.
1. Galileo deduced that the acceleration due to gravity (when no other forces interfere) would be the
same for all objects regardless of mass.
a) Explain how this is possible when the two masses have different weights.
b) If you take a sheet of paper from a book and crumple the paper into a small ball it will fall at
essentially the same rate as the book itself. However, if you don’t crumple the paper it will
soon attain a constant velocity and it will stop accelerating. Explain.
c) Convince yourself that given the mass of the Earth and its radius g=9.8 m/s2.
d) Determine the altitude above the surface of the Earth where the acceleration is half that at the
surface.
e) Look up the mass of the Moon and its radius and convince yourself that the acceleration due to
gravity on the surface of the Moon is 1/6 as much as on the surface of the Earth.
a) A larger mass requires more force to accelerate it. In the case of gravity the larger mass has a larger
weight in just the right proportion to give it the same acceleration as a smaller mass with corresponding
smaller weight. Another way to explain this is to say that mass is proportional to both the inertia of an
object and to its gravitational pull (mass has inertial and gravitational properties in equal proportion). You
can also see this most simply when you equate FG to “mg”, the mass of the body drops out and “g” emerges
as a constant for all masses.
b) Air resistance will act in opposition to the gravitational force. As the speed increases, so does air
resistance until it equals and cancels the weight of the object at which point the object moves with constant
(terminal) velocity.
c) Done in class.
d) Fg=mgsurface /2 at a distance (1.41Re) from the center of the earth, or (0.4Re) above the surface.
e) Get the necessary values from the table and plug into the formula for “g”.
2. The distance between the Earth and the Moon (center to center) is about 60 earth radii and it takes the
Moon 27.3 days to complete a revolution. The Sun is 23,500 earth radii from the center of the Earth, and it
takes 365.25 days for the Earth to complete one revolution around the Sun. The Sun’s mass is 333,000
times greater than that of the Earth. Illustration is not to scale.
60RE
Moon
Sun
Earth
23,500RE
a) Determine the acceleration of the Moon toward the Earth due to gravity. Compare that to the
acceleration of the Earth towards the Sun.
b) Show that the acceleration of the Moon toward the Earth due to gravity is equal to the centripetal
acceleration of the Moon in orbit around the Earth. Newton submitted this agreement as proof that his Law
of Gravity was correct. Explain Newton’s reasoning.
c) When the Moon is between the Earth and the Sun (as in a solar eclipse), the Moon is pulled by both in
opposite directions. Compare the force of attraction toward the Sun to the force of attraction toward the
Earth.
d) Given the answer to (b), explain why the Sun doesn’t just pull the Moon away from the earth.
a) gm/e=9.8/602= 0.00273 m/s2; ge/s=GMs/(re/s)2= 0.0059 m/s2….
b) ac= v2/rm/e = 0.0027 m/s2 , which compares well with answer in (a). Newton’s argument was that this
equality supported his theory that there was a universal force of gravity pulling the Moon toward the Earth
whose strength decreased with distance inversely squared.
c) You need to look up the Sun-Moon distance (same as the Sun-Earth distance of 1.5x1011 m) and the MonEarth distance (3.6x108 m), also the masses of Sun and Earth to plug into the ratio [Fsun-moon/Fearth-moon]. I
got about 189.
d) As you can see in (c) the Sun pulls the Moon harder than the Earth but the Moon isn’t pulled away
because it is orbiting the Sun along with the Earth as they both go around the Sun in a year.
3. Consider a satellite in (circular) orbit around the earth. Determine the speed necessary for the satellite to
remain in orbit at the following locations:
v
a) Near the surface of the Earth.
b) At an arbitrary distance “r” from the center of the Earth.
Is this speed slower or faster than at the surface?
c) In a geo-synchronous orbit around the equator. This is an orbit in
which the satellite has the same period as the Earth’s daily rotation about its axis (24 hours).
d) Explain why geo-synchronous orbits are useful for communication satellites.
a) v=(gRE)1/2 = 7908 m/s
b) The orbital speed depends inversely on distance [v= (2x107)(r)-1/2], so the farther away the slower the
satellite.
c) Since rgeosync =6.6RE, you can calculate v =2π(6.6 RE)/T(1day in sec)=3070 m/s
d) These satellites remain at rest relative to the surface of the earth so one can aim a satellite dish directly
at them without worrying about re-aiming it.
4. A generation before Newton a mathematician named Johannes Kepler discovered three laws of
planetary motion from painstaking analysis of vast amounts of observational data accumulated by the great
astronomer Tycho Brahe. One of Newton’s triumphs was to be able to show that his theories of gravitation
and motion predicted Kepler’s laws.
Kepler’s 3rd law states that “the radius of an orbit cubed divided by the period squared is a
constant” for all the planets that orbit the Sun (R3/T2=Kepler’s constant). Derive this law from
Newtonian principles.
b) A convenient unit of distance for the solar system is the “astronomical unit” (AU) which is the
orbital distance between the Earth and the Sun (1AU= 1.50 x 1011 m). Astronomers prefer this unit
for obvious reasons as well as preferring the “earth-year” (365.25 days) to measure time. Determine
Kepler’s constant (K) in these units. As a comparison, determine this constant in “mks” units.
c) Mars’ orbit around the Sun takes 1.9 (earth) years. Determine the distance between Mars and the
Sun compared to the Earth-Sun distance. (Hint: Use AU and earth-year units.)
d) Kepler’s 3rd Law can be applied to comets as well as planets. But because comets have very
eccentric orbits the “R” in the equation is an average between the furthest distance from the Sun that
the comet attains (apehelion) and the closest distance that the comet attains (perihelion). Comet
Halley approaches the Sun to within 0.570 AU, and its orbital period is 75.6 years. What is the
a)
furthest that Halley goes in its orbit? (Hint: Compare Halley’s orbit to the Earth.)
0.57 AU
Earth
Halley’s comet
at perihelion
Halley’s comet
at apehelion
1 AU
e) Satellites that orbit the Earth also obey Kepler’s 3rd Law but the constant would be different.
Explain why and determine the value of Kepler’s constant for the Earth.
a) This was done in class. Start with Fc=FG. Kepler’s constant “K”= GM/42=r3/T2.
b) K=1 in AU and earth-years. Ksun= 3.36 x 1018 m3/s2.
c) Class example, Radius of Mars orbit=1.53 AU
d) Comparing Halley’s comet to Earth give an average radius for Halley’s comet of 17.9 AU. This makes
the apehelion (furthest) distance = 35.2 AU, when averaged with the given perihelion (closest ) distance of
0.570 AU. In contrast, Pluto’s average radius of orbit = 39.4 AU.
e) Kearth= 10.1 x 1012 m3/s2. The “M” in Kepler’s law refers to the large body or “mother” that attracts
the satellites.
5. Many of the stars we observe are actually double star systems. Two stars close to each other will go
around their mutual center of mass. It was believed these double-star systems would not support stable
planetary orbits, but recently astronomers were surprised to find a planet orbiting a double star system!
a) Consider two stars of equal mass M a distance D apart.
The center of mass is the point in between. Determine
Center of mass
the speed of the stars in orbit and the period of the rotation.
M
Note that the each star travels in a circle of radius D/2.
M
b) If you were standing on one of the stars how would you
D
describe the motion of the other star? Could you tell from
observations alone that you were “moving”?
c) In the general case of this problem the stars have different masses,
Center of mass
M
M and m. The center of mass is closer to the larger star and
rM
m
its distance from the larger mass is given by the formula
rM =[Dm/(m+M)]. Determine the speed of each star in its orbit.
D
d) Draw circles describing the orbits of the two stars.
e) Show that the stars have the same period, that is, that they take the same time to complete a full
rotation.
a) Start by equating the centripetal force to the gravitational force (Fc=FG). But realize that the distance D is
twice the radius of orbit of these stars. Answer: v =(GM/2D)1/2.
b) You would see the other star go around you with a radius of orbit equal to D. In this case it would be very
hard for you to conclude who is really moving and who is standing still.
c) Start the same way as in (a). Each star however orbits in a different size circle around the center of mass.
For example, the larger star orbits with radius rM. Answers: vM =[Gm2/(m+M)D] 1/2; vm =[GM2/(m+M)D] 1/2.
d) Both stars circle the center of mass, but the larger mass has a smaller radius and the smaller has a larger
radius.
e) The period T=2πr/v. Whichever star you choose, you get the same answer: T=[4π2D3/G(M+m)]1/2. The
two masses remain the same distance apart as they rotate.
6. The earth’s daily rotation (24 hours) about its axis has an effect on our “apparent weight”. In this
problem we will compare a person standing on the North Pole to a person standing on the equator.
a) Determine the centripetal acceleration of a person
standing on the equator due to the daily rotation of the Earth
about its axis. Is there a similar centripetal
acceleration on the North Pole? Explain.
b) Draw a “fbd” for a person standing on the North Pole. How does
24 hrs
the normal force here compare to the person’s true weight (mg)?
c) Draw a “fbd” for a person standing on the equator. How does the
normal force here compare to the person’s true weight (mg)?
d) Where do you “feel lighter”, on the equator or on the pole? Determine the ratio of the apparent
weight at the equator to the apparent weight at the pole.
e) How long would the day have to be to cause a person to lose contact with the ground due to the
daily rotation of the Earth? Would this happen at the equator or at the pole?
a) Determine the tangential speed at the equator by dividing the circumference of the Earth by the 24 hour
day. Then use the formula for centripetal acceleration, you get ac= 0.0338 m/s2. There is no acceleration at
the poles because the person is at the center of the rotation and has no speed.
b) The only forces are weight toward the center and normal away from the center (N-mg=Fc). At the pole
the forces are equal and the person feels his normal “mg” weight.
c) Because of the centripetal acceleration the normal is less than the weight.
d) You would feel lighter than your true weight at the equator because N is less. At the poles you feel your
normal weight. Ratio: Normalequator/Normal pole=0.997=99.7% (not a big difference)
e) This could only happen at the equator. N=zero, so mg=mv2/Re and v=(g Re)1/2. For the speed at the
equator to be this fast the “day” would have to be only 1.4 hours long.
7. You should feel confident that you understand the difference between mass, weight, and “apparent
weight”.
a) Define these terms unambiguously and give examples illustrating the difference.
b) The space shuttle is on a near-earth orbit yet the astronauts float around. Do they float around
because they have less gravitational force than on the surface of the Earth? Explain.
c) Someday we may build and live inside space stations far from Earth. In these stations the pull of
gravity would be nearly zero and the problem arises of how to provide a “feeling of weight” for the
inhabitants so that their muscles and bones do not weaken. One technique for creating such
“artificial gravity” is to build a rotating space station (the classic science fiction film “2001-A
Space Odyssey” demonstrates this idea very nicely). The station would consist of a spinning
cylinder where people would walk on the inside wall. With the appropriate combination of radius
and rotational speed the feeling of standing on a planet can be approximated.
If the radius of the cylinder were 5 m, how fast would the station have
to spin for people to “feel” like they have their “Earth-weight”?
d) If the person were to run in the direction of the rotation of the
cylinder, how would he feel? How would he feel if were to run
in the opposite direction as the rotation of the cylinder?
Person inside
cylinder
a) Go over the definitions…
b) No, the shuttle is actually near the Earth in its orbit where the pull of gravity is still strong. The
astronauts float because they are in “free-fall” along with the shuttle in orbit. Both astronaut and ship
are accelerating at “g” but are not making contact with each other, hence the feeling of
“weightlessness”.
c) Here the normal force between the person and the inner surface of the cylinder provides the “feeling
of weight” as is also the centripetal force. You want that normal to equal the weight of the person on
Earth. Hence, N=mv2/R=mg. Solve for v.
d) If you run in the same direction as the rotation of the station, your speed increases relative to the
center and you will feel heavier because “N” would be larger. Running in the opposite direction will
decrease your speed relative to the center so you feel lighter.
8. Kepler’s 3rd Law can be used to determine the unknown mass of a body.
If you know the period and position of a satellite, that information can be used to determine the mass of
the body that the satellite orbits.
a) Show that you can determine the mass of the earth from knowing the radius and period of the
Moon’s orbit.
b) Use the distance of the Earth from the Sun and the year to determine the mass of the Sun.
c) Io, a satellite of Jupiter, has an orbital period of 1.77 days and an orbital radius of 4.22 x 105 km.
Determine the mass of Jupiter.
a) Kepler’s 3rd Law is: r3/T2 =GM/4π2. If you know the radius of orbit “r” and the period of orbit “T” of
an earthly satellite, then you can calculate “M” which would be the mass of the Earth. Astronomers
often use this method to find the mass of planets by observing their satellites. Mearth =5.98 x 1024 kg.
b) Using the same method, Msun is about 2 x 1030 kg. You need to look up the Sun-Earth distance (1AU=
1.50 x 1011 m).
c) Mass of Jupiter is about 2 x 1027 kg.
9. Newton’s Law of Gravity gives the force of attraction between two point masses. For extended bodies
the question arises of what is the proper distance “d” between the bodies. For spherically symmetric bodies
the answer is simply “center-to-center”. This is not as obvious as it seems at first, and indeed it takes
calculus to prove it. Since calculus did not exist in Newton’s time, Newton had to “invent” the calculus in
order to develop his physical theories, which makes him not only one the greatest scientists of all time but
also one the best mathematicians as well.
For non-spherical geometries the “distance between objects” implied in the law of gravitation is NOT the
distance between the centers of the two bodies, as you can demonstrate through integration. Take for
example the force of gravitational attraction between a point mass “m” and a one-dimensional uniform rod
of mass M and length L. In this problem you will derive an expression for the gravitational force between
them by breaking up the rod into “infinitesimal pieces” (dM) and integrating the “infinitesimal forces”
between them and “m”:
L
D
m
M
First consider:
M
x
x
a) Start by highlighting a piece of the rod, M, at an arbitrary distance “x” from the end. Write an
expression for the gravitational force (F) between this piece alone and the point mass “m”. Don’t
forget to properly include the distance D in your expression.
b) Express M in terms of x, L and M.
c) Now change the “’s” into differential “d’s” as you consider the limit. You now have an integral in
terms of the variable “x”. Integrate this function from x=0 to x=L and you get the answer.
d) Is it correct to assume the rod behaves as if all its mass were concentrated at its center?
a) F= (GmM)/(D+x)2
b) Since the “linear density” of the rod is uniform, M/x=M/LM=(x)M/L.
c) dF = GmdxM/L)/(D+x)2. After separating the variable “x” from the constants, you can integrate the
function [dx)/(D+x)2] from 0 to L. The final answer is: Fg =GmM/D(D+L)
d) No, if the rod behaved as if all its mass were concentrated at its center, then Fg = GmM/(D+L/2)2,
which is not the correct value as determined by integration and would underestimate the force of
gravity.
Challenge problem
10. Because the Earth is (more or less) spherically symmetric and a person is small enough (compared to
the Earth) to be considered a point mass, we can determine our weight by measuring our distance from
the center of the Earth and not from the surface. But this doesn’t work if we go below the Earth’s
surface because different parts of the Earth pull you in opposite directions and the effect of gravity
partially cancels out. In this problem assume that you can pass through the center of the Earth along a
a) What should be the force of gravity on your body if you were at the center of the Earth? Justify your
answer. Assume a constant diameter, and that the density of the Earth is uniform throughout.
b) It can be proven than when you are inside the Earth a distance “r” from the center, the gravitational
effect of all the mass of the Earth outside the radius “r” cancels out. Taking this statement as a
“given”, determine the force of attraction on your body from the mass of Earth enclosed within the
radius “r”. This would be your “weight” inside the Earth and you should be able to see that it is
proportional to “r”.
c) Sketch a graph of your weight vs. distance from the center of the Earth from r=0 to r=4RE. Use your
graph to determine the two locations where your weight will be half the value at the surface.
d) If you are really good with math and want a challenge
you can try to prove the assertion made in (b). But you
Person
inside Earth
can also give a conceptual argument that supports the
assertion.
r
RE
Portion of mass
that cancels out.
a) Force
of gravity would be zero because the Earth is pulling symmetrically in all directions.
b) The mass within the radius “r” (let’s call it Mr) can be calculated from the density (assumed constant)
and volume. You can show that Mr=ME(r/RE)3. Now you can use the Law of Gravity to get: Fat r= Gm Mr /r2.
After substituting for M, the answer is Fat r= Gm ME r/RE3. You weight decreases linearly with “r” as you
penetrate into the Earth.
c) The two locations are at r= RE /2 and at r=1.41 RE.
d) If you think of the Earth as being made up of spherical shells like the layers of an onion, you can consider
one layer from the inside. Opposite sections on the shells exert equal and opposite forces on any point
inside (there are few bits of mass close to the point in question and many bits far away). This basic idea can
be shown to be mathematically rigorous. Some more advanced physics or math textbooks will contain this
proof.