Question

What is the period of a satellite in a circular orbit just above the surface of the Moon? The Moon’s mass is 7.36 × 1022 kilograms and its radius is
1.738 × 106 meters.


A.
1.94 × 104 seconds
B.
1.30 × 104 seconds
C.
6.50 × 103 seconds
D.
3.24 × 103 seconds
E.
1.62 × 103 seconds

Answer 1

Answer:

C. 6.50 x 10^3

Explanation:

The satellite revolves around moon due to centripetal force which is provided by gravity  in this context so we can equate gravity and centripetal force.

                  $mrw^{2} = \frac{GMm}{R^{2} }$

where, m is mass of satellite

           M is mass of moon

           ω is angular velocity

           G is Universal gravitational constant

            R is the radius of moon

from the above equation we get ω as,

                          $w=\sqrt{\frac{GM}{R^{3} } }$

                               $T = \frac{2\pi }{w}$

                                   $T= 2\pi \sqrt{\frac{R^{3} }{GM} }$

      here M = 7.36 X 10^22

               R = 1.738 X 10^6

               G =6.67 X 10^-11

              $T= 2X3.14 \sqrt{\frac{(1.738 X 10^6)^{3} }{6.67 X 10^-11 X 7.36 X 10^22} }$

                      T= 6.50 X 10^3 seconds