**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 : **

Figure %: alignment of Mercury, Venus and the Sun.

F_{s} = = 5.54×10^{22} |

The distance between Mercury and Venus is given by

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?

**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

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

= = |

Rearranging and then substituting the expression we found for

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

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

Take a Study Break!