Question

calculate the speed of revolution of a polar satellite close to the surface of the earth . Given , R=6400 km, g=9.8 m/s^2​

Answer 1

Answer:

Speed of revolutions of the satellite is given as

v = 7919.6 m/s

Explanation:

As we know that gravitational force of Earth on the satellite is net centripetal force on it

So we will have

$F_c = mg$

here we know that

$F_c = \frac{mv^2}{R}$

now we have

$\frac{mv^2}{R} = mg$

$v = \sqrt{Rg}$

$v = \sqrt{(6400 \times 10^3)(9.8)}$

$v = 7919.6 m/s$

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Topic : Orbital Speed

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Answer 2

Answer:

hope it helps you

Explanation:

As we know that gravitational force of Earth on the satellite is net centripetal force on it

So we will have

F_c = mgF

c

=mg

here we know that

F_c = \frac{mv^2}{R}F

c

=

R

mv

2

now we have

\frac{mv^2}{R} = mg

R

mv

2

=mg

v = \sqrt{Rg}v=

Rg

v = \sqrt{(6400 \times 10^3)(9.8)}v=

(6400×10

3

)(9.8)

v = 7919.6 m/sv=7919.6m/s