Problem :
What is the force exerted by Big Ben on the Empire State building? Assume that
Big Ben has a mass of
10^{8}
kilograms and the Empire State building
10^{9}
kilograms. The distance between them is about 5000 kilometers and Big Ben is
due east of the Empire State building.
The direction of the force clearly attracts the Empire State towards Big Ben.
So the direction is a vector pointing due east from New York. The magnitude is
given by Newton's Law:
F = = = 2.67×10^{7}
N


Clearly, the gravitational force is negligibly small, even for quite large
objects.
Problem :
What is the gravitational force that the sun exerts on the earth? The
earth on
the sun? In what direction do these act? (
M
_{e} = 5.98×10^{24}
and
M
_{s} = 1.99×10^{30}
and the earthsun distance is
150×10^{9}
meters).
First, consider the directions. The force acts along the direction such that it
attracts each body radially
along a line towards their common center of mass. For most practical purposes,
this means a line
connecting the center of the sun to the center of the earth. The magnitude of
both forces is the same, as we
would expect from Newton's Third
Law, and they act in
opposite directions, both
attracting each other mutually. The magnitude is given by:
F = = = 3.53×10^{22}


Problem :
Figure %: alignment of Mercury, Venus and the Sun.
If Mercury, Venus and the sun are aligned in a right triangle, as shown, then
calculate the vector sum of the
forces on Venus due to both Mercury and the Sun. What is the direction and
magnitude of the resulting
force? (SunVenus distance
r
_{v} = 108×10^{9}
meters, SunMercury distance
r
_{m} = 57.6×10^{9}
meters, mass of Sun
M
_{s} = 1.99×10^{30}
kilograms, mass of Mercury
M
_{m} = 3.3×10^{23}
kilograms, mass of Venus
M
_{v} = 4.87×10^{24}
kilograms).
The magnitude of the force on Venus due to the sun is given by:
F
_{s} = = 5.54×10^{22}


The distance between Mercury and Venus is given by
r
_{mv} = = 1.08×10^{11}
meters. The magnitude of the force from Mercury, then, is:
F
_{m} = = 9.19×10^{15}


The directions of these forces are along the lines connecting the planets. If
the size of the forces was comparable, we would have to resolve each vector
force into components perpendicular and parallel to some direction, and then sum
these components in order to find the final direction of the force. In this
case however, the force due to the sun is more than a million times greater than
the force due to Mercury, and so the net force is very well approximated by the
magnitude and direction of the force due to the sun.
Problem :
It is possible to simulate "weightless" conditions by flying a plane in an arc
such that the centripetal acceleration exactly cancels the acceleration due to
gravity. Such a plane was used by NASA when training astronauts. What would be
the required speed at the top of an arc of radius 1000 metres?
We require an acceleration that exactly cancels that due to gravity  that is,
exactly 9.8 m/sec
^{2}
. Centripetal acceleration is given by
a
_{c} =
.
We have been given
r = 1000
meters, so
v = =
99
m/s.
Problem :
Show using Newton's Universal Law of Gravitation that the period of orbit of
a binary star system is given by:
T
^{2} =


Where
m
_{1}
and
m
_{2}
are the masses of the respective stars and
d
is the
distance between them. Notice that we derived the same result in a problem in
the previous section, using the reduced mass and Kepler's Third Law.
Consider the center of mass of the binary star system to be at the origin.
Since the planets must be on opposite sides of their center of mass, it must be
true that
m
_{1}
r
_{1}  m
_{2}
r
_{2} = 0
where
r
_{1}
and
r
_{2}
are the radii of orbit.
Since
r
_{2} + r
_{1} = dâá’r
_{2} = d  r
_{1}
, we can write
m
_{1}
r
_{1} = m
_{2}
r
_{2} = m
_{2}(d  r
_{1})
.
Rearranging, we can solve for
r
_{1}
:
r
_{1} =
d
.
Now the force acting between the two masses is given by Newton's Law:
F =


We can proceed as we did in deriving Kepler's Third Law from Newton's Law, and
say that this force must be equal to the centripetal force acting on
m
_{1}
:
Rearranging and then substituting the expression we found for
r
_{1}
, we have:
Which is the same result we derived from Kepler's Third Law.