Answered: The mass of Kepler is what fraction of…

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Question

there was a new planet discovered in 2017 names Kepler, it is about 300 light years from Earth. Compared to the expo planet BEE106 that was discovered in 2014, Kepler has 3 times less radius. The acceleration due to gravity on BEE106 is 3 times the gravitational acceleration on Kepler. The mass of Kepler is what fraction of the mass of BEE106.

Expert Solution

Step 1

Let an object of mass $m$ lies on the surface of a planet of mass $M$ of radius $R$ . The force of attraction between the planet and the object is given by Newton's Law of Universal Gravitation

$F = \frac{G M m}{R^{2}}$

$G$ is the universal gravitational constant. This force is equal to the weight of the object on the planet. Let the acceleration due to the gravity is $g$ then the weight of the object on the planet is

$F = m g$

Thus we get

$m g = \frac{G M m}{R^{2}} \Rightarrow g = \frac{G M}{R^{2}}$

Step 2

Let the mass and radius of Kepler be $M$ and $R$ respectively and the mass and radius of planet BEE106 be $M'$ and $R'$ .

It is given that the radius of Kepler is three times less than that of BEE106. Thus

$R' = 3 R$

Let the acceleration due to gravity on Kepler be $g$ and that of BEE106 be $g'$ , it is given in the question that

$g' = 3 g$

Therefore the acceleration due to gravity on Kepler

$g = \frac{G M}{R^{2}}$

Acceleration due to gravity on BEE106

$g' = \frac{G M'}{R'^{2}} = \frac{G M'}{{(3 R)}^{2}}$

Taking the ratio

$\frac{g}{g'} = \frac{\frac{G M}{R^{2}}}{\frac{G M'}{{(3 R)}^{2}}} \Rightarrow \frac{g}{3 g} = \frac{M}{M'} \frac{9 R^{2}}{R^{2}} \Rightarrow \frac{M}{M'} = \frac{1}{3 \times 9} = \frac{1}{27} \Rightarrow M = \frac{1}{27} M'$

This shows that the mass of Kepler is $\frac{1}{27}$ times the mass of BEE106.

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