Question 159: Gravitation

Question

The time period of a satellite in a circular orbit of radius R is T. The period of another satellite in a circular orbit of radius 9R is :

Accepted Answer

Kepler's Third Law states that the square of the period of a satellite's orbit is proportional to the cube of the semi-major axis of its orbit. This relationship can be written as:
$$T^2 \propto r^3$$
where:
$T$ is the orbital period
$r$ is the radius of the circular orbit (which serves as the semi-major axis in this case)
Considering two satellites, one with period $T$ and radius $R$, and another with unknown period $T'$ and radius $9R$, we can form an equation:
$$\frac{{T'}^2}{T^2} = \frac{(9R)^3}{R^3}$$
This simplifies to:
$$\frac{{T'}^2}{T^2} = 729$$
Taking the square root of both sides, we get:
$$T' = T 	imes \sqrt{729} = T 	imes 27$$
So, the period of another satellite in a circular orbit of radius $9R$ is $27T$.

Answer

9 T

Answer

27 T

Answer

12 T

Answer

3 T

Gravitation

The time period of a satellite in a circular orbit of radius R is T. The period of another satellite in a circular orbit of radius 9R is :

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