**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.

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 earth-sun distance is 150×10^{9} meters).

F = = = 3.53×10^{22} |

**Problem : **

*r*

_{v}= 108×10

^{9}meters, Sun-Mercury 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?

^{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}=

*d*âá’

*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:

T^{2} = = d^{3} = |

Which is the same result we derived from Kepler's Third Law.