Question 74: Gravitation

Question

If a satellite orbiting the Earth is 9 times closer to the Earth than the Moon, what is the time period of rotation of the satellite? Given rotational time period of Moon $=27$ days and gravitational attraction between the satellite and the moon is neglected.

Accepted Answer

To find the time period of the satellite, we can use KeplerÃ¢ÂÂs third law, which states that the square of the orbital period is proportional to the cube of the orbital radius. Mathematically, this is expressed as:
$$\frac{T^2}{R^3} = 	ext{constant}$$
HereÃ¢ÂÂs how we solve the problem step by step:
Let:
The Moon's distance from Earth be $$R_{	ext{moon}}$$.
The Moon's orbital period be $$T_{	ext{moon}} = 27$$ days.
The satellite's orbital radius be $$R_{	ext{sat}} = \frac{R_{	ext{moon}}}{9}$$ (since it is 9 times closer).
According to Kepler's law for both orbits, we have:
$$\frac{T_{	ext{sat}}^2}{R_{	ext{sat}}^3} = \frac{T_{	ext{moon}}^2}{R_{	ext{moon}}^3}$$
Substitute $$R_{	ext{sat}}$$:
$$\frac{T_{	ext{sat}}^2}{\left(\frac{R_{	ext{moon}}}{9}ight)^3} = \frac{27^2}{R_{	ext{moon}}^3}$$
Simplify the left side by noting that:
$$\left(\frac{R_{	ext{moon}}}{9}ight)^3 = \frac{R_{	ext{moon}}^3}{9^3} = \frac{R_{	ext{moon}}^3}{729}$$
So the equation becomes:
$$\frac{T_{	ext{sat}}^2}{\frac{R_{	ext{moon}}^3}{729}} = \frac{27^2}{R_{	ext{moon}}^3}$$
Multiply both sides by $$\frac{R_{	ext{moon}}^3}{729}$$:
$$T_{	ext{sat}}^2 = 27^2 	imes \frac{1}{729}$$
Recognize that $$27^2 = 729$$:
$$T_{	ext{sat}}^2 = \frac{729}{729} = 1$$
Take the square root:
$$T_{	ext{sat}} = 1 	ext{ day}$$
Thus, the satelliteÃ¢ÂÂs orbital period is 1 day, which corresponds to
Option D: 1 day
.

Answer

3 days

Answer

27 days

Answer

81 days

Answer

1 day

Gravitation

If a satellite orbiting the Earth is 9 times closer to the Earth than the Moon, what is the time period of rotation of the satellite? Given rotational time period of Moon =27 days and gravitational attraction between the satellite and the moon is neglected.

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