Download Exploration of Space Lecture B

Exploration of Space Lecture B
In the previous lesson, we saw the great contributions the Kepler
made to the understanding of how planets and other objects move around the
sun. Kepler realized that the sun somehow controlled the motion of the
planets, but he was unable to explain how it did this. It was left to Newton,
fifty years later, to provide a satisfactory answer. In this lesson we will see
how Newton developed his Law of Universal Gravitation.
Newton graduated from Cambridge University in 1665 after majoring
in mathematics. In the few years after this, Newton did some of his best
thinking. Legend has it that while watching an apple fall, Newton began to
wonder if the force that accelerated the fall of the apple might also be the
one responsible for maintaining the path of the moon’s orbit. The moon is
moving around Earth in a circular orbit and thus has centripetal acceleration.
This acceleration is caused by the gravitational pull between the moon and
Earth. The amount of this acceleration due to gravity was proportional to the
gravitational force between Earth and the moon. By using his knowledge of
circular motion and Kepler’s Third Law, Newton deduced that the
gravitational force was inversely proportional to the square of the distance
from Earth to the moon.
Newton’s reasoning was similar to the following. From our work on
circular motion, we know that the centripetal force required to keep an
object moving in a circle is
From Kepler’s Law,
By substituting this value for T2 into the equation for force, we obtain
Since the expression (42K) is a constant, the force of gravitational
attraction on a planet is directly proportional to the mass of the planet and
inversely proportional to the square of its distance from the sun. We saw in
the previous lesson that Kepler’s constant is the same for any object circling
the sun, but it would have a different and smaller value for a satellite of
Earth. Newton reasoned that this is because the mass of Earth is smaller than
the mass of the sun. In other words, Kepler’s constant is proportional to the
mass of the object exerting the force of attraction at the centre of the orbit.
So the expression 42K is proportional to mc where mc refers to the mass of
the central body (like the sun or Earth). We can thus rewrite this expression
as 42K = G mc. The value G is referred to as a universal gravitational
constant. Henry Cavendish first experimentally measured it in 1798.
Newton’s remarkable extension was to say that if this equation for force
applies to the sun and its planets, and to Earth and its moon, then it should
apply to any body in the universe that has mass. We can summarize
Newton’s Law of Universal Gravitation as
Any two bodies attract each other with forces proportional to the mass
of each and inversely proportional to the square of the distance
between them.
where G = 6.67 x 10-11 Nm2/kg2
Example
Two basketballs of mass 0.65 kg are placed 1.0 m apart. What is the
force of gravitational attraction between them?
Solution
F = (6.67 x 10-11 Nm2/kg2)(0.65 kg)(0.65 kg)/(1.0)2 = 2.8 x 10-11 N
Example
The moon has a mass of 7.34 x 1022 kg and the mass of the earth is
5.98 x 1024 kg. The force of gravitational attraction between the moon and
the earth is 2.00 x 1020 N. What must be the distance between the moon and
the earth?
Solution
Example
Proportional reasoning:
Two objects of identical mass are found to attract each other with a
gravitational force of 10 N. What happens to this 10 N if:
a)
b)
c)
d)
the mass of one of the objects is doubled
both the masses are doubled
the distance between them is doubled
the mass of one of the objects is doubled and the distance between
them is doubled
Solution
Look at the formula to guide your thinking.
a) the force will be multiplied by two (2 x 10 N = 20 N)
b) the force will be multiplied by four (4 x 10 N = 40 N)
c) think of it as 1/R2, and R = 2. Therefore the force will be
multiplied by ¼ (1/4 x 10 N = 2.5 N)
d) the force will be multiplied first by 2, but then multiplied by ¼ (2 x
10 N x ¼ = 5 N)