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:
[tex] F_{c} = F_{G} [/tex]
[tex]\frac{mv^{2}}{r + h} = \frac{GMm}{(r + h)^{2}}[/tex]
[tex] v = \sqrt{\frac{gr^{2}}{r+h} [/tex]
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
[tex]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[/tex]
b) The period of its revolution is:
[tex] 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 [/tex]
c) The gravitational force acting on it is given by:
[tex] F = \frac{GMm}{(r + h)^{2}} [/tex]
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
[tex] 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 [/tex]
I hope it helps you!
