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#### Dhamnekar Winod

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- Nov 17, 2018

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- Thread starter Dhamnekar Winod
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- Nov 17, 2018

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- Mar 5, 2012

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We are given the angular velocity $\omega = 7\cdot 10^{-5}\,rad/s$ and the mass $M=6\cdot 10^{24}\,kg$.

To achieve a free fall of $0\,m/s^2$ at radius $r$ we need that the centripetal acceleration is equal to the acceleration due to gravity,

Note that $v=\omega r$, so the centripetal acceleration is $\frac{v^2}{r}=\omega^2 r$.

The acceleration due to gravity is $\frac{GM}{r^2}$, where $G=6.67\cdot 10^{-11}$ is the gravitational constant (leaving out the unit while assuming SI units).

So:

$$\omega^2 r = \frac{GM}{r^2}$$

Solve for $r$.

To achieve a free fall of $0\,m/s^2$ at radius $r$ we need that the centripetal acceleration is equal to the acceleration due to gravity,

Note that $v=\omega r$, so the centripetal acceleration is $\frac{v^2}{r}=\omega^2 r$.

The acceleration due to gravity is $\frac{GM}{r^2}$, where $G=6.67\cdot 10^{-11}$ is the gravitational constant (leaving out the unit while assuming SI units).

So:

$$\omega^2 r = \frac{GM}{r^2}$$

Solve for $r$.

Last edited:

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- #3

- Nov 17, 2018

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Hi,We are given the angular velocity $\omega = 7\cdot 10^{-5}\,rad/s$ and the mass $M=6\cdot 10^{24}\,kg$.

To achieve a free fall of $0\,m/s^2$ at radius $r$ we need that the centripetal acceleration is equal to the acceleration due to gravity,

Note that $v=\omega r$, so the centripetal acceleration is $\frac{v^2}{r}=\omega^2 r$.

The acceleration due to gravity is $\frac{GM}{r^2}$, where $G=6.67\cdot 10^{-11}$ is the gravitational constant (leaving out the unit while assuming SI units).

So:

$$\omega^2 r = \frac{GM}{r^2}$$

Solve for $r$.

So, we get $r^3 =8.172587755e22m^3/rad^2$ So,$r=43396349.43332m/\sqrt[3]{rad^2}$. Is this answer correct?

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- Mar 5, 2012

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I get the same answer.Hi,

So, we get $r^3 =8.172587755e22m^3/rad^2$ So,$r=43396349.43332m/\sqrt[3]{rad^2}$. Is this answer correct?

Do note that $rad$ is not an actual physical unit, but it's a ratio. When we multiply the angular velocity (rad/s) with the radius (m), the rad unit is effectively eliminated and we get m/s.

So properly we have $r=4.3\cdot 10^7\,m$.

It means that answer 2 should be the correct answer.

Admittedly it's a bit strange that it is given as $4.4\cdot 10^7\,m$ instead of $4.3\cdot 10^7\,m$.

Since we're talking about earth, perhaps they used a mass and angular velocity with a higher precision than the ones given in the problem statement.

EDIT: Hmm... in that case we would actually get $r=4.2\cdot 10^7\,m$, so that can't be it after all.

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