An object of mass m orbits a star with mass M , where M m . In general the bound or bits are elliptical, which means the kinetic energy of the object T and the gravitational potential energy U may vary. For bound orbits, the total energy E is negative. If the orbit is circular, then T and U are both constant.Show that for circular orbits T : U : E = 1 : -2 : -1.

Question

An object of mass m orbits a star with mass M , where M >> m . In general the bound or bits are elliptical, which means the kinetic energy of the object T and the gravitational potential energy U may vary. For bound orbits, the total energy E is negative. If the orbit is circular, then T and U are both constant.Show that for circular orbits T : U : E = 1 : -2 : -1.

Damon · Answer

well, let's take zero of U at infinity so at radius r from star

U = - G M m/r

the force holding it in orbit is mass * centripetal acceleration
G M m/r^2 = m v^2/r
so
m v^2 = G M m /r Oh, my look at that
so

(1/2)m v^2 = -(1/2) U = Ke

Now the total is the sum of those which is
-(1/2) U + U = (1/2)U

the kinetic energy = (1/2) m v^2