The time period of revolution of moon around the earth is 28 days and radius of its orbit is `4xx10^(5)` km. If `G=6.67xx10^(-11) Nm^(2)//kg^(2)` then

Question

The time period of revolution of moon around the earth is 28 days and radius of its orbit is `4xx10^(5)` km. If `G=6.67xx10^(-11) Nm^(2)//kg^(2)` then

The time period of revolution of moon around the earth is 28 days and radius of its orbit is `4xx10^(5)` km. If `G=6.67xx10^(-11) Nm^(2)//kg^(2)` then find the mass of the earth.

Monjur Alahi

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2025-01-04 04:42

1 Answers

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Salim Ahmed Kawsar · Accepted Answer · 2023-02-05T15:29:48+06:00

Salim Ahmed Kawsar

Correct Answer - `6.47xx10^(24)` kg

8 views Answered 2023-02-05 15:29