Question:
The period of revolution of an earth's satellite close to surface of earth is 90 min. The time period of another satellite in an orbit at a distance of four times the radius of earth from its surface will be
The period of revolution of an earth's satellite close to surface of earth is 90 min. The time period of another satellite in an orbit at a distance of four times the radius of earth from its surface will be
Updated On: Jul 7, 2022
- $90\sqrt{9} $ min
- 270 min
- 720 min
- 360 min
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The Correct Option is C
Solution and Explanation
From Kepler's law
$T^2\propto R^3$
or $T\propto R^{3/2}$
$\frac{T'}{T}=\left(\frac{R'}{R}\right)^{3/2}$
or $\frac{T'}{T}=\left(\frac{4R}{R}\right)^{3/2}$
= $(4)^{3/2}=(2^2)^{3/2}=2^3=8$
= $T' =8T = 8\times90$
= 720 min
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Concepts Used:
Escape Speed
Escape speed is the minimum speed, which is required by the object to escape from the gravitational influence of a plannet. Escape speed for Earth’s surface is 11,186 m/sec.
The formula for escape speed is given below:
ve = (2GM / r)1/2
where ,
ve = Escape Velocity
G = Universal Gravitational Constant
M = Mass of the body to be escaped from
r = Distance from the centre of the mass