# How would the speed of the Earth have to change in the following case? If our Sun were seventeen times as massive as it is, how many times faster or slower should the Earth move in order to remain in the same orbit? For an object revolving around another object, there has to be a force of attraction between the two. This is the gravitational attraction between the Earth and the Sun.

The force of attraction is G*Me*Ms/r^2

The centripetal force is Me*V^2 / r.

The two are equal at equilibrium, so we have G*Me*Ms/r^2 = Me*V^2 / r.

=> Vi^2 = (G*Me*Ms/r^2)/ (Me/r)

=> Vi^2 = (G*Ms/r)

Now, if the mass of the sun became 17 times its present mass(I have assumed you meant an increase in mass and not one in the diameter) and the orbit of the Earth around the Sun has to remain the same,

G*Me*17*Ms/r^2 = Me*Vf^2 / r

=> Vf^2 = (G*17*Ms/r)

=> Vf^2 = 17*Vi^2

=> Vf = sqrt 17 * Vi

So the velocity of the Earth would have to be sqrt 17 times its present velocity if the orbit has to remain the same.

Approved by eNotes Editorial Team