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.

The correct answer is  "c".  it is "c" because melting is an endothermic change, and endothermic get's energy from surroundings.