Question

A 782-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. 9.82278e7 Incorrect: Your answer is incorrect. Your response differs significantly from the correct answer. Rework your solution from the beginning and check each step carefully. m/s (b) Find the period of its revolution. h (c) Find the gravitational force acting on it.

Answer 1

Answer:

a) v = 5.59x10³ m/s

b) T = 4 h

c) F = 1.92x10³ N

Explanation:

a) We can find the satellite's orbital speed by equating the centripetal force and the gravitation force as follows:

$F_{c} = F_{G}$

$\frac{mv^{2}}{r + h} = \frac{GMm}{(r + h)^{2}}$

$v = \sqrt{\frac{gr^{2}}{r+h}$          

Where:

g is the gravity = 9.81 m/s²        

r: is the Earth's radius = 6371 km

h: is the satellite's height = r = 6371 km      

$v = \sqrt{\frac{gr^{2}}{2r}} = \sqrt{\frac{gr}{2}} = \sqrt{\frac{9.81 m/s^{2}*6.371 \cdot 10^{6} m}{2}} = 5.59 \cdot 10^{3} m/s$                                      

b) The period of its revolution is:

$T = \frac{2\pi}{\omega} = \frac{2\pi (r + h)}{v} = \frac{2\pi (2*6.371 \cdot 10^{6} m)}{5.59 \cdot 10^{3} m/s} = 14322.07 s = 4 h$

c) The gravitational force acting on it is given by:

$F = \frac{GMm}{(r + h)^{2}}$

Where:

M is the Earth's mass =  5.97x10²⁴ kg    

m is the satellite's mass = 782 kg

G is the gravitational constant = 6.67x10⁻¹¹ Nm²kg⁻²

$F = \frac{GMm}{(r + h)^{2}} = \frac{6.67 \cdot 10^{-11} Nm^{2}kg^{-2}*5.97 \cdot 10^{24} kg*782 kg}{(2*6.371 \cdot 10^{6} m)^{2}} = 1.92 \cdot 10^{3} N$

I hope it helps you!