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

h = 1.014 x
`10^(8)` m

Step-by-step 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{(GM)/(R + h) }`

Since

V = (R+h)ω

Substituting in the above equation

(R+h)ω =
`\sqrt{(GM)/(R + h) }`

Squaring on both sides

{(R+h)ω}² =
`{(GM)/(R + h) }`

(R+h)³ =
`{(GM)/(w^(2))}`

Solving for h

h =
`\sqrt[3]{{(GM)/(w^(2))}}` - R

Substituting the values in the above equation

h =
`\sqrt[3]{(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