GRAVITY LAB
This procedure takes you through the law of gravitation to Black Holes in a series of
problems you are supposed to solve. Enjoy!
1. Gravitational law of attraction: The gravitational force Fg between two spherically
symmetric objects of mass m1 and m2 is proportional to the product of their masses,
and falls off inversely as the square of the distance r between their centers of masses:
F =
GN m1 m2
r2
where GN is the Gravitational (or Newton’s) Constant.
(a) Find the physical dimensions of GN (in terms of [M], [L] & [T]).
(b) What is the value for GN ? Include proper SI units.
(c) Two objects are moved twice their distance apart. The gravitational force between
them diminishes by a factor of
. They are moved back to a
third of their original distance apart. Now the gravitational force between them
increases by a factor of
.
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2. The gravitational force between a body of mass m1 and the Earth, m2 = ME , is called
the weight of the body m1 . But the weight of the body on Earth’s surface is also given
.
by F = m1 gE , where the value with units of gE =
By equating the two expressions for force, find gE in terms of GN , ME , RE . (Thus,
knowing the size of the Earth and the acceleration due to gravity, it is possible to
“weigh” the Earth— i.e., find its mass. This is how we know the Earth’s mass!)
As a balloon ascends into the atmosphere, it
we go from Earth’s surface, the smaller the value of
weight, because the higher
.
3. The average density ρE of the Earth is defined to be its mass ME divided by its total
volume. Find an expression relating gE to average density and radius of the Earth.
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Given that gE = 9.8 m/s2 , RE = 6.4 × 106 m, and using the value of GN from Q.1,
part(b), find the average density of the Earth.
A planet half the radius of the Earth, RP = 21 RE , is determined to have (using a
pendulum) a gP = 4gE . How does its average density compare to that of Earth’s?
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4. The gravitational force of the Earth causes a Gravitational potential energy on an
object of mass m a distance r from the center, where r ≥ RE , given by
Ug (r) = −
GN ME m
.
r
Derive the (approximate) expression for potential energy at a height h above Earth’s
surface by evaluating the difference between the gravitational PE at the two places;
that is, show that when h << RE , then
Ug (RE + h) − Ug (RE ) = −
GN ME m
GN ME m
− (−
) ≈ mgE h.
RE + h
RE
Hint: you would want to employ the Binomial Theorem Approximation, (1 + x)n ≈
1 + nx, provided that |x| << 1.
When you use ∆Ug = mgE h in physics problems, you are assuming that height h is
compared with RE , and that for this height, gE is
.
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5. The total energy of a body is the sum of its gravitational potential energy and its kinetic
energy. Write an expression for the total energy of a satellite orbiting the Earth in a
circular orbit at a distance R from Earth’s center, where R ≥ RE , assuming its orbit
speed to be v.
Find the satellite’s orbit speed v as follows: equate the centripetal force needed for
going in a circular orbit to the gravitational force between the satellite and the Earth.
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Suppose the orbit is completed in time T ; find an expression relating T and R (Kepler’s
third law of motion!).
Geostationary Orbits: When the period of orbit of a satellite coincides with the earth’s
rotational period, then it is said to be in an orbit that’s stationary with respect to
earth. (Such a satellite will remain in a fixed location in the sky.) Find the relationship
between R and RE for this to happen. Evaluate the height above the earth’s surface
in kilometers.
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Ground-hugging Orbits: Find the period of a satellite which barely clears the ground
as it orbits the Earth. Determine the speed of such a satellite, and give the answer in
both m/s and mph.
Substitute back the expression for satellite speed v into the expression for its total
energy. You should find that the total energy is always negative, a universal property
of gravitating systems. The orbit is called a “bound state” orbit. The satellite will be
trapped in this orbit unless enough extra kinetic energy is given to it to escape.
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6. Escape velocity. If enough KE is given to an object on Earth’s surface to equal its
gravitational PE with respect to the Earth, then the body would escape into space
without ever falling back. The total energy of the body is zero, and at infinity, its
speed will eventually be zero as well. Find escape speed vEe in terms of GN , ME and
RE .
Find an expression for the velocity of escape, vE e in terms of gE at the surface, & RE .
Does the escape velocity depend on the mass of the ejected object? Circle: Y / N
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Knowing gE & RE compute vE e the escape speed for Earth, in both km/s and miles
per second. How long at this speed will an object take to go across continental USA?
The Moon is such that RM = 0.27 × RE & gM = 61 × gE . Find the ratio of the escape
velocity from the Moon, vM e to that from the Earth vEe .
Knowing vE e from the previous question, find vM e in miles/sec.
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7. An object that is such that even light v e = c = 3 × 108 m/s cannot escape from its
grasp is called a Black Hole, abbreviated henceforth “BH”.
Take a beam of light of mass m and equate its escape KE to its gravitational PE on
the BH surface. This gives a relationship between RBH & MBH that involves in it
constants c & G. The spherical surface of radius RBH is called the event horizon of
the BH.
is its size.
The more massive a black hole, the
Find RBH for (i) the Earth (ii) the Sun, (iii) the entire Universe. In each case, look
up the data for mass MBH of the object as it exists now. Note that the BH radius of
the Universe is also the radius of the observable Universe! We live in a big black hole.
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8. Find a relation between density of BH material and the mass of a black hole.
is its BH material.
The more massive a black hole, the
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Evaluate solar BH density. Use M = 2 × 10
previous page.
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kg, and RBH as determined in the
For fun, let us find the volume of a rectangular lake with the same mass of water as
the mass of one teaspoonful (= 5 cm3 ) of solar black hole material. If the lake is one
mile by one mile square, find its depth. Water density is ρW = 1.0 gm/cm3 .
Show that the surface (i.e., event horizon radius) value of the acceleration due to gravity
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of a BH is given by gBH = 2RcBH . Evaluate the radius of a BH for which this is equal to
gE = 9.8 m/s2 . Find also its mass in terms of M . The existence of such supermassive
black holes proves that statements such as “as you enter a black hole, you will be torn
apart” are not always true. That only applies if the black holes are small.
The End. Submit at the rate of 10 extra credit points per page!
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