# If the mass of earth were 2 times the present mass,the mass of the moon were half the present mass and the moon were revolving round the earth at the earth at the same present distance, the time period of revolution of the moon would be(in days)?

May 22, 2018

$\approx 19$ days

#### Explanation:

We can derive an expression for the time period of revolution by assuming that the moon is moving in a circular orbit around the earth.

The necessary centripetal acceleration is provided by the force of gravity and thus we have

${m}_{m} {\omega}^{2} r = G \frac{{m}_{e} {m}_{m}}{r} ^ 2 \implies$

${\omega}^{2} = G {m}_{e} / {r}^{3} \implies$

$T = \frac{2 \pi}{\omega} = \frac{2 \pi}{\sqrt{G {m}_{e}}} {r}^{\frac{3}{2}}$

Thus, under the conditions of the problem the period will decrease by a factor of $\sqrt{2}$ to become $\frac{27}{\sqrt{2}}$ days $\approx 19$ days.
The text passage contains a physics problem that involves a mathematical question and provides an answer to it, along with an explanation of the solution.
single_question_with_answer
Given a text passage, determine:
    1. If it contains a mathematical question (including word problems or calculations)
    2. If an answer to that mathematical question is provided in the text

First, describe the content structure you found in 1-2 sentences.

Then provide your analysis in the following format:
TYPE: <one of the following values>
- NO_QUESTION
- SINGLE_QUESTION
- SINGLE_QUESTION_WITH_ANSWER
- MULTIPLE_QUESTIONS
- MULTIPLE_QUESTIONS_WITH_ANSWERS
system
# If the mass of earth were 2 times the present mass,the mass of the moon were half the present mass and the moon were revolving round the earth at the earth at the same present distance, the time period of revolution of the moon would be(in days)?

May 22, 2018

$\approx 19$ days

#### Explanation:

We can derive an expression for the time period of revolution by assuming that the moon is moving in a circular orbit around the earth.

The necessary centripetal acceleration is provided by the force of gravity and thus we have

${m}_{m} {\omega}^{2} r = G \frac{{m}_{e} {m}_{m}}{r} ^ 2 \implies$

${\omega}^{2} = G {m}_{e} / {r}^{3} \implies$

$T = \frac{2 \pi}{\omega} = \frac{2 \pi}{\sqrt{G {m}_{e}}} {r}^{\frac{3}{2}}$

Thus, under the conditions of the problem the period will decrease by a factor of $\sqrt{2}$ to become $\frac{27}{\sqrt{2}}$ days $\approx 19$ days.
user
The text passage contains a physics problem that involves a mathematical question and provides an answer to it, along with an explanation of the solution.

TYPE: SINGLE_QUESTION_WITH_ANSWER