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If the earth contracts to half its radius, what would be the length of the day?
Answer
469.8k+ views
Hint: We know that the earth is a part of the solar system. And we also know that currently, we have 24 hours in a day. From Kepler's law of planetary motion, we know that the time period of revolution of the planets is proportional to the distance of the planet from the Sun.
Formula Used:
$F_{C}=mR\omega^{2}$
Complete step-by-step solution
We know that the earth is the 3rd planet from the sun in the solar system and it is the only planet, which can sustain life out of the 8 planets. We know that there is a gravitational force between the sun and the earth.
From newton’s law of gravitation we know that the gravitational force between the sun and the earth is given as $F=\dfrac{Gm_{1}m_{2}}{R^{2}}$ where $G$ is the gravitational constant, $m_{1}$ and $m_{2}$ are the masses of the sun and the earth respectively, which are separated at a distance $R$ from each other.
Since the planets are constantly moving on the circular path, we can say that they experience centripetal force towards the center of the sun, this force is given as $F_{C}=mR\omega^{2}$ where $\omega$ is their angular velocity.
Given that the radius of the earth is contracted to half its radius. Then for them to remain in its orbital the two forces must be equal.
$\implies mR\omega^{2}=m\dfrac{R}{2}\omega^{2}$
If the mass remains the same, we have,
$\implies R\left(\dfrac{1}{T}\right)^{2}=\dfrac{R}{2}\left(\dfrac{1}{T\prime}\right)^{2}$
$\implies \left(\dfrac{1}{T}\right)^{2}=\dfrac{1}{2}\left(\dfrac{1}{T\prime}\right)^{2}$
We know that $T=24h$
$\implies \left(\dfrac{1}{24}\right)^{2}=\dfrac{1}{2}\left(\dfrac{1}{T\prime}\right)^{2}$
$\implies \left(\dfrac{1}{24}\right)=\sqrt{\dfrac{1}{2}}\left(\dfrac{1}{T\prime}\right)$
$\implies T\prime=\dfrac{24}{\sqrt2}=19.97h$
$\therefore T\prime=19.97h$ if the earth contracts to half its radius.
Note: We also know that the earth rotates around its own axis and revolves around the sun in a specific orbit. The earth stays in its orbit due to the gravitational force between the earth and the sun. Also, we know that there are 24 hours in a day.
Formula Used:
$F_{C}=mR\omega^{2}$
Complete step-by-step solution
We know that the earth is the 3rd planet from the sun in the solar system and it is the only planet, which can sustain life out of the 8 planets. We know that there is a gravitational force between the sun and the earth.
From newton’s law of gravitation we know that the gravitational force between the sun and the earth is given as $F=\dfrac{Gm_{1}m_{2}}{R^{2}}$ where $G$ is the gravitational constant, $m_{1}$ and $m_{2}$ are the masses of the sun and the earth respectively, which are separated at a distance $R$ from each other.
Since the planets are constantly moving on the circular path, we can say that they experience centripetal force towards the center of the sun, this force is given as $F_{C}=mR\omega^{2}$ where $\omega$ is their angular velocity.
Given that the radius of the earth is contracted to half its radius. Then for them to remain in its orbital the two forces must be equal.
$\implies mR\omega^{2}=m\dfrac{R}{2}\omega^{2}$
If the mass remains the same, we have,
$\implies R\left(\dfrac{1}{T}\right)^{2}=\dfrac{R}{2}\left(\dfrac{1}{T\prime}\right)^{2}$
$\implies \left(\dfrac{1}{T}\right)^{2}=\dfrac{1}{2}\left(\dfrac{1}{T\prime}\right)^{2}$
We know that $T=24h$
$\implies \left(\dfrac{1}{24}\right)^{2}=\dfrac{1}{2}\left(\dfrac{1}{T\prime}\right)^{2}$
$\implies \left(\dfrac{1}{24}\right)=\sqrt{\dfrac{1}{2}}\left(\dfrac{1}{T\prime}\right)$
$\implies T\prime=\dfrac{24}{\sqrt2}=19.97h$
$\therefore T\prime=19.97h$ if the earth contracts to half its radius.
Note: We also know that the earth rotates around its own axis and revolves around the sun in a specific orbit. The earth stays in its orbit due to the gravitational force between the earth and the sun. Also, we know that there are 24 hours in a day.
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