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:

Clearly, the gravitational force is negligibly small, even for quite large objects.

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:

The magnitude of the force on Venus due to the sun is given by:

The distance between Mercury and Venus is given by r _mv = = 1.08×10¹¹ meters. The magnitude of the force from Mercury, then, is:

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.

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.

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:

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.