If a satellite orbiting the Earth is 9 times closer to the Earth… — Solution | JEE Physics

MCQEasyJEE 2025Kepler's Laws of Planetary Motion

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 = $27 \, \text{days}$ and gravitational attraction between the satellite and the moon is neglected.

A
$1 \, \text{day}$
B
$81 \, \text{days}$
C
$27 \, \text{days}$
D
$3 \, \text{days}$

Answer

Correct answer:A

Step-by-step solution

Standard Method

Given: The Moon has time period $T_1 = 27 \, \text{days}$ and the satellite is at distance $R_2 = \frac{R_1}{9}$ from Earth, where $R_1$ is the Moon's distance.

Find: The time period $T_2$ of the satellite.

For two bodies revolving around the same central body, Kepler's third law gives

\frac{T_1^2}{R_1^3} = \frac{T_2^2}{R_2^3}

Substitute $R_2 = \frac{R_1}{9}$ :

\frac{T_1^2}{R_1^3} = \frac{T_2^2}{\left(\frac{R_1}{9}\right)^3}

So,

\frac{T_1^2}{R_1^3} = \frac{T_2^2}{R_1^3/729}

Hence,

T_2^2 = T_1^2 \cdot \frac{1}{729}

Using $T_1 = 27 \, \text{days}$ ,

T_2 = 27 \cdot \left(\frac{1}{9}\right)^{3/2}

Now,

\left(\frac{1}{9}\right)^{3/2} = \frac{1}{27}

Therefore,

T_2 = 27 \cdot \frac{1}{27} = 1 \, \text{day}

Therefore, the time period of rotation of the satellite is $1 \, \text{day}$ . The correct option is A.

Using direct proportionality

Given: $T \propto r^{3/2}$ , Moon's time period is $27 \, \text{days}$ , and the satellite is $9$ times closer than the Moon.

Find: The satellite's time period.

From Kepler's third law,

T \propto r^{3/2}

So,

\frac{T_{\text{satellite}}}{T_{\text{moon}}} = \left(\frac{r_{\text{satellite}}}{r_{\text{moon}}}\right)^{3/2}

Using

r_{\text{satellite}} = \frac{r_{\text{moon}}}{9}

we get

T_{\text{satellite}} = 27 \cdot \left(\frac{1}{9}\right)^{3/2} = 27 \cdot \frac{1}{27} = 1 \, \text{day}

This works because the same central mass Earth is governing both motions, so only the orbital radius ratio is needed. Therefore, the correct option is A.

Note on discrepancy in extracted working

The solution contains an intermediate line equivalent to $T_2^2 = 27^2 \times 729$ , which is inconsistent with Kepler's third law for a smaller orbital radius. A smaller radius must give a smaller time period. The later conclusion $T_2 = 1 \, \text{day}$ is correct, and it matches the second approach as well as the option marked on the page.

Common mistakes

Using $T \propto r^3$ instead of $T^2 \propto r^3$ is incorrect. Kepler's third law relates the square of time period to the cube of orbital radius. First write $T \propto r^{3/2}$ , then compare the two bodies.
Interpreting " $9$ times closer" as $r_{\text{satellite}} = r_{\text{moon}} - 9$ is wrong. It means the satellite's distance is $\frac{1}{9}$ of the Moon's distance. Use $r_{\text{satellite}} = \frac{r_{\text{moon}}}{9}$ .
Assuming a smaller orbital radius gives a larger time period is conceptually wrong here. For satellites around the same central body, decreasing $r$ decreases $T$ . Check whether your final answer is less than $27 \, \text{days}$ .

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