Question

What is the period of revolution polar satellite orbiting close to the surface of the earth (R=6400km,g=9.8/s2

Answer 1

Answer:

Explanation: Given: Radius of earth = R = 6400 km = 6.4 x 106 m, Time period = T = 27.3 days = 27.3 x 24 x 60 x 60, g = 9.8 m/s6.

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Answer 2

Answer:

The time period of the revolution polar satellite orbiting close to the surface of the earth is 83.7 minutes.

Explanation:

The time period of revolution of the polar satellite orbiting close to the surface of the earth is given as,

$T=2\pi\sqrt{\frac{d}{g} }$          (1)

Where,

T=time period of revolution of the satellite

d=distance of the satellite from the surface of the earth

g=acceleration due to gravity

From the question we have,

The radius of the earth(R)=6400km=6.4×10⁶m  (1km=1000m)

g=9.8m/s²

Since the satellite orbiting close to the surface of the earth d will be equal to "R". So, we can write equation (1) as,

$T=2\pi\sqrt{\frac{R}{g} }$         (2)

By substituting the required values in equation (2) we get;

$T=2\pi\sqrt{\frac{6.4\times 10^6}{9.8} }$

$T=2\pi\times 10^3\times 0.80$

$T=5.024\times 10^3 second$

$T=83.7minutes$

Hence, the time period of the revolution polar satellite orbiting close to the surface of the earth is 83.7 minutes.