Question

A "synchronous" satellite is put in orbit about a planet to always remain above the same point on the planet’s equator. The planet rotates once every 10.2 h, has a mass of 2.2 × 1027 kg and a radius of 6.99 × 107 m. Given: G = 6.67 × 10−11 N m2 /kg2 . Calculate how far above the planet’s surface the satellite must be

Answer 1

Answer:

h = 1.014 x $10^{8}$ m

Explanation:

Given that,

The planet rotates once every 10.2 h

Therefore, in one hour it would take 1/10.2 rev/h

The rotational speed of the planet

                               = 0.098 rev/h

Converting into seconds

                                = 2.72 x $10^{-5}$ rev/s

Converting to radians, the rotational angular velocity is

                             ω  = 1.709 x $10^{-4}$ rad/s

The mass of the planet, M = 2.2  x $10^{27}$ Kg

Radius of the planet,      R  = 6.99 x $10^{7}$  m

Gravitational constant,   G  = 6.67 x $10^{-11}$  Nm²/Kg²

To find the height 'h' of the synchronous orbit above the planet surface,

The orbital velocity of the planet is given by the expression

                                      V  = $\sqrt{\frac{GM}{R + h} }$

Since

                                       V  = (R+h)ω

Substituting in the above equation

                                (R+h)ω = $\sqrt{\frac{GM}{R + h} }$

Squaring on both sides

                                {(R+h)ω}² = ${\frac{GM}{R + h} }$

                                (R+h)³ =  ${\frac{GM}{w^{2}}}$

Solving for h

                                    h  =  $\sqrt[3]{{\frac{GM}{w^{2}}}}$  - R

Substituting the values in the above equation

                 h =  $\sqrt[3]{\frac{6.67X10^{-11}X2.2X10^{27} }{(1.709X10 ^{-4})^{2} }}$  m

                                   h  = 1.014 x $10^{8}$ m

So, the height of the synchronous orbit above the surface is 1.014 x $10^{8}$ m