Name___________________________________________________ Date_______________________________
Investigation 6
Gravitation
According the Newton’s Universal Law of Gravitation, all masses exert attractive forces on all other
masses. This force is called the gravitational force. This force is proportional to the product of the masses
(mass 1) x (mass 2) , and is inversely proportional to the distances between them squared (1/distance2). It is
written as:
(mass 1) x (mass 2)
gravitations force 
,
(distance) 2
or as an equation:
gravitational force  G x
(mass 1) x (mass 2)
(distance) 2
,
where G is called the universal gravitational constant (= 6.67 x 10 -11 N m2/kg2).
A planet with a mass M and its moon with a mass m are a distance d
apart.
1.
2.
Does the larger planet exert a greater force on the moon, or does
the moon exert a greater force on the planet? Explain.
M
m
d
By how much does the force of attraction change (e.g., doubles, halves, quadrupled, etc.) when the
following occurs separately?
a.
the mass of the planet doubles to 2M?
b.
the mass of the planet doubles and the mass of the moon triples?
c.
the distance between the planet and the moon decreases to 1/3 of the original distance?
d.
the mass of the planet triples and the distance between the planet and the moon doubles?
These are the units for weight and mass in the three common systems of units.
Weight
Mass
Unit in the SI System of
Units
Unit in the British
Engineering System of
Units
Unit in the cgs System of
Units
Newtons (N)
pounds (lb)
dynes
kilograms (kg)
slugs
grams (g)
2.
What weighs more, one kilogram of lead or one kilogram of feathers? (Do you know why?)
3.
a.
How much do you weigh? Express your weight in pounds.
b.
Write your weight in Newtons. In order to do this, you must multiply your weight in pounds
by 4.45.
c.
Now calculate your mass in kilograms. Remember that weight = mass x g, and that mass =
(weight/g).
d.
The force the earth exerts on you as calculated by using Newton’s Universal Law of Gravitation
m m
is G 1 2 2 , where G = 6.67 x 10-11 N m2/kg2, m1 = mass of earth = 5.98 x 1024 kg, m2 = your mass
d
in kg, and d = radius of the earth = 6.378 x 106 m. Should the value of this force be the same as
the value of your weight found in part b? Explain – do not calculate.
e.
4.
The gravitational force on objects on the surface of the moon is approximately 1/6 of the
gravitational force the earth exerts on the same objects.
1)
What is the value of your mass when you are on the moon? Express your result in
kilograms.
2)
How much do you weigh on the moon? Express your result in pounds and Newtons.
A physics student weighs 600 N at the surface of the earth. Find her weight at the following
positions:
a.
a distance above the surface of the earth equal to the radius of the earth.
b.
a distance above the surface of the earth equal to four times the radius of the earth.
c.
at the center of the earth.
5.
Let’s take a look at what happens to the gravitational force on a student who weighs 600 N as the
radius of the earth is reduced, but the mass remains the same (the density of the earth will increase).
This may give you some idea of how a black hole results from the gravitational collapse of a star.
The gravitational force on a person at the surface of the earth (weight) is
M  x m person
, where mperson is the person’s mass. This is called an
Force or Weight   G x earth
(R earth ) 2
“inverse square” force. This means that when the distance is doubled, the force is reduced by a factor
of 2 squared, or four. That is, the force is ¼ of its original value. And when the distance is halved,
the force is increased by a factor of 2 squared, or four. That is, the force is four times greater. If the
distance is tripled, then the force is decreased by a factor of 9 (3 squared) – one-ninth of the original
value. And when the distance is reduced to a third, then the force is increased by a factor of 9 – nine
times larger. And so on.


R

So, if the radius of the earth is reduced to half of its value  earth  , then the 600 N student will
2


increase in weight to 4 x 600 = 2400 N. If the radius of the earth is reduced to one-third of its original
R

value,  earth  , then the student will weigh 9 x 600 = 5400 N, and so on.
3


a.
Suppose the radius of the earth is reduced to one-tenth (1/10) of its original value, then how
much will the student weigh at the surface of the earth? (Remember that the mass of the earth
isn’t changing.)
b.
When the radius of the earth is reduced to about one millimeter (remember, all of the mass of
the earth is contained within that 1 mm radius sphere), the earth theoretically becomes a black
R earth


 . How much does the student now
hole. In this case, the radius of the earth is 
10,000,000
,000


weigh? (Obviously, this can’t happen, but it is fun to speculate.)