A 606-kg satellite is in a circular orbit about Earth at a height above Earth eq

Question

A 606-kg satellite is in a circular orbit about Earth at a height above Earth equal to Earth's mean radius.
(a) Find the satellite's orbital speed.
(b) Find the period of its revolution.

(c) Find the gravitational force acting on it.
A 606-kg satellite is in a circular orbit about Earth at a height above Earth equal to Earth's mean radius.
(a) Find the satellite's orbital speed.
(b) Find the period of its revolution.

(c) Find the gravitational force acting on it.
A 606-kg satellite is in a circular orbit about Earth at a height above Earth equal to Earth's mean radius.
(a) Find the satellite's orbital speed.
(b) Find the period of its revolution.

(c) Find the gravitational force acting on it.

Explanation / Answer

Here ,

a)

radius of earth , R = 6371 km

Now ,

satellite's orbital speed = sqrt(G * M/r)

satellite's orbital speed = sqrt(6.673 *10^-11 * 5.98 *10^24/(2 * 6371 * 1000))

satellite's orbital speed = 5596.2 m/s

the satellite's orbital speed is 5596.2 m/s

b)

let the period of its revolution = 2*pi*r/v

period of its revolution = 2 * pi * 2 * 6371*10^3/(5596.2)

period of its revolution = 14303.5 s

the period of its revolution is 14303.5 s

c)

gravitaional force = G * M *m/d^2

gravitaional force = 6.673 *10^-11 * 5.98 *10^24*606/(2 * 6371 * 1000)^2

gravitaional force = 1489.4 N

the gravitaional force on the satellite is 1489.4 N

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