Universal Gravitation Assignment KEY
Level 1
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Tychonic system – a planetary model in which Sun and Moon revolve around Earth, but the other planets
revolve around the Sun
Kepler’s laws – three empirical relationships that describe the motion of planets
Law of universal gravitation – the force of gravity between any two objects is proportional to the product of
their masses and inversely proportional to the square of the distance between their centers
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Geostationary orbit – the orbit of a satellite around Earth’s equator, which gives the satellite the appearance of
hovering over the same spot on Earth’s surface at all times
Microgravity – the condition of an object that experiences apparent weightlessness (or is in free fall)
Perturbation – deviation of a body in orbit from its regular path, caused by the presence of one or more other
bodies
Inverse square law – the relationship in which the force between two objects is inversely proportional to the
square of the distance that separates the centers of the objects
Action at a distance – the description of the force between two objects not in contact
Field – a region in space that influences a mass, charge, or magnet placed in the region
Gravitational field intensity – the quotient of the gravitational force and the magnitude of the test mass at a
given point in the field
Gravitational field lines – imaginary directed lines that indicate the path a tiny test mass would take if free to
move in the gravitational field; these lines radiate inward toward the mass that generates them
Level 2
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Level 3
• The graph below of force (x 10-8 N) vs. 1/r2 (x 103 m-2) was developed from data obtained by measuring the
force of attraction between two identical 7.00 x 102 g masses at various separation distances.
a) Draw the line of best fit. Calculate the
35
slope of the line of best fit below.
SEE GRAPH; 3.18 x 10-11 N m2
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b) Approximate the value of the universal
)
gravitational constant, G, by using the
N25
result from a).
8
^20
FG = G(m1*m2)/r2
0
2
-11
2
1
Thus FG/r = G(m1*m2) = 3.18 x 10 N m
x
-11
2
( 15
To find G divide 3.18 x 10 N m by (0.7
e
2
-11
2
2
kg) = 6.5 x 10 N m /kg
c
r
o10
c) An identical experiment is carried out on
F
the surface of the Moon, where the
acceleration due to gravity is 1/6 that on
5
the surface of the Earth. How does this
0
affect the value of G? Using the
0
2
4
6
8
10
12
principles of physics, explain your
reasoning.
1/r^2 (x 10^3 m^-2)
“G” would not be affected since it is a
universal constant anywhere
d) On a popular science TV show, the host described the orbiting space shuttle astronauts as being
“weightless” because they are in a “zero gravity” environment. Using field concepts, explain why the
phrase “zero gravity” should not be used.
“Zero gravity” does not exist since the force of gravity never actually reaches zero, only approaches zero
as the sdistance between the shuttle and the Earth increases.
•
Four 7.5-kg spheres are located at the corners of a square of side length 0.60 m. Calculate the magnitude
and direction of the net gravitational field at the center of this configuration. SHOW ALL WORK.
The distance from each sphere to the center of the square is 0.424 m. Thus, using g = G*m/r2, g = 2.78 x
10-9 N/kg for each sphere. But for every attractive gravitational field, there is an opposing attractive
gravitational field from a sphere at the diagonal corner so the field intensities add up to give a net 0
N/kg.