. ms – , R E = . × m, the distance to the moon R = . × m and the time period of the moon’s revolution is .
days. Obtain the mass of the Earth M E in two different ways. Answer From Eq. ( .
× kg. The moon is a satellite of the Earth. From the derivation of Kepler’s third law [see Eq. ( .
. - . kg Both methods yield almost the same answer, the difference between them being less than %. Example .
Express the constant k of Eq. ( . ) in days and kilometres. Given k = – s m – .
The moon is at a distance of . × km from the earth. Obtain its time-period of revolution in days. Answer Given k = – s m – km = .
× – d km – Using Eq. ( . ) and the given value of k, the time period of the moon is T = ( . × - )( .
× ) T = . d Note that Eq. ( . ) also holds for elliptical orbits if we replace ( R E + h ) by the semi-major axis of the ellipse.
The earth will then be at one of the foci of this ellipse. . ENERGY OF AN ORBITING SATELLITE Using Eq. ( .
), the kinetic energy of the satellite in a circular orbit with speed v is K E m v ( Gm M , ( . ) Considering gravitational potential energy at infinity to be zero, the potential energy at distance (R e +h) from the centre of the earth is G m M P E ( . ) The K.E is positive whereas the P.E is negative. However, in magnitude the K.E.