Sol. The satellite is moving outside of the earth's atmosphere so that the only force acting on it is the gravitational attraction of the earth. With the mass and radius of the earth expressed by m e and R , respectively, the gravitational law of Eq. gives F = Gmm e /r 2 = gR 2 m/r 2 when the substitution. Gm e = gR 2 is made for the surface values F = mg and r = R. The work done by F is due only to the radial component of motion along the line of action of F and is negative for increasing r .

U 1–2 = –

The work-energy equation U 1-2 = T gives

mgR 2 = m (v 2 2 – v 1 2 ) v 2 2 = v 1 2 + 2gR 2

Substituting the numerical values gives

v 2 2 = + 2(9.81) [(6371) (10 3 )] 2

= 69.44(10 6 ) – 10.72(10 6 ) = 58.73 (10 6 ) (m/s) 2

v 2 = 7663 m/s or v 2 = 7663(3.6) = 27590 km/h Ans .

Helpful Hints:

(i) Note that the result is independent of the mass of the satellite.

(ii) Consult Table D/2, appendix D, to find the radius R of the earth.