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The moon takes about 27.3 days to revolve around the earth in a nearly circular orbit of radius...

Question:

The moon takes about 27.3 days to revolve around the earth in a nearly circular orbit of radius {eq}3.84\times10^5 km {/eq}. Calculate the mass of the earth from these data.

Time Period and Mass:


The mathematical expression for the orbital time period of a planet (from the third law of Kepler) is shown below:

{eq}\displaystyle T=2\pi \sqrt{\frac{R^3}{GM}} {/eq}, where,

  • R is the radius of a circular orbit measured in meters.
  • M is the mass of larger planet or earth.
  • G is the universal gravitational constant and it is equal to {eq}\rm 6.67\times 10^{-11}\ N\cdot m^2/kg^2 {/eq}.

Answer and Explanation:


Given information:

  • The orbital time period of the moon is {eq}T=\rm 27.3\ days {/eq}.
  • The radius of the circular orbit is {eq}R=\rm 3.84\times10^5\ km=\rm 3.84\times10^5\times10^3\ m {/eq}.

Simplifying the value of time period in seconds, we get:

{eq}\begin{align*} \displaystyle T&=\rm 27.3\ days \times \frac{24\ hours}{1\ day}\\ \displaystyle T&=\rm 655.2\ hours\\ \displaystyle T&=\rm 655.2\ hours\times \frac{3600\ seconds}{1\ hour}\\ \displaystyle T&=\rm 655.2\times 3600\ seconds\\ \displaystyle T&=\rm 2358720\ seconds\\ \end{align*} {/eq}

Here, we have:

{eq}\displaystyle T=2\pi \sqrt{\frac{R^3}{GM}} {/eq}

Squaring both sides of the above expression and simplifying it for the required mass, we have:

{eq}\begin{align*} \displaystyle T^2&=\left(2\pi \sqrt{\frac{R^3}{GM}}\right)^2\\ \displaystyle T^2&=4\pi^2\left( \frac{R^3}{GM}\right)\\ \displaystyle M&=4\pi^2\left( \frac{R^3}{GT^2}\right)\\ \end{align*} {/eq}

Substitute the required values in the above expression and simplify it.

{eq}\begin{align*} \displaystyle M&=4\pi^2\left( \frac{(\rm 3.84\times10^5\times10^3\ m)^3}{(\rm 6.67\times 10^{-11}\ N\cdot m^2/kg^2)(\rm 2358720\ seconds)^2}\right)\\ &=\displaystyle 4\pi^2\left( \frac{\rm 56.623104\times10^{24}\ m^3}{\rm 371.089\ m^3/ kg}\right)\\ &=\boxed{\rm 6.024\times10^{24}\ kg} \end{align*} {/eq}

Thus, the mass of the earth is {eq}6.024\times10^{24} {/eq} kilograms.


Learn more about this topic:

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Kepler's Three Laws of Planetary Motion

from Basics of Astronomy

Chapter 22 / Lesson 12
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