Calculate the velocity of a 1650 kg satellite that is in a circular orbit of 4.2 x 106m above the surface of a planet, which has a radius 3.95 x 105 m and the period of the orbit is 2.0 hours.

A. 1.4 x 107m/h

B. 1.24 x 106m/h

C. 1.3 x 107m/h

D. 1.45 x 107m/h

only answer if you know it

First of all go through the pic. Now from (ii) equation, M = 4 (pie^2) r^3 / GT^2. M = 4 x pie^2 x (4595000) ^3 / 6.67x (10^ - 24) x (7200) ^2 [ T = 2hrs = 2 x 60x60 = 7200 seconds]. We get M = 1.10770842 x 10^24 kg. Put value of M in equation (i). Therefore, V = square root[ (6.67x10^-11 x 1.10770842 x 10^24) / 4595000) ]. We get, V = 4009.893956 meters/second. To convert this into metres / hour, 4009.893956 x 3600 [since 1hr = 3600seconds]. We get V = 14435618.24 meters/hour i. e 1.44 m/h. So option A is the correct answer.