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I understand that for momentum to be conserved, if I throw a tennis ball (or kick a football) the Earth must move in the opposite direction to the ball.

Obviously this is an infinitesimally small amount, but how could how calculate much would the Earth's rotation speeds up or slows down (depending on the direction of the kick/throw) by this action? Let's say that a football weighs 0.45kg and I kick it to a speed of 25 metres per second, can I work out the effect this has on the Earth?

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    $\begingroup$ Sure. Momentum is conserved, so $m_{ball}\Delta v_{ball}=m_{Earth}\Delta v_{Earth}$. Given that Earth has a mass of $6\times 10^{24}kg$ nothing noticeable will happen. $\endgroup$ – CuriousOne Jun 11 '16 at 21:29
  • $\begingroup$ Yes you can work it out using conservation of angular momentum. $\endgroup$ – sammy gerbil Jun 11 '16 at 21:41
  • $\begingroup$ Similar Mankind vs Earth questions: physics.stackexchange.com/q/70732/2451, physics.stackexchange.com/q/56245/2451 and links therein. $\endgroup$ – Qmechanic Jun 11 '16 at 22:16
  • $\begingroup$ Thanks for the pointers. It is a thought experiment, as I realise that nothing tangible would happen. Of course you could argue that, statistically speaking, given the number of people on the planet kicking balls (or doing similar momentum-related activities) at any given instant of time, their directions would equal each other out anyway. $\endgroup$ – Dr Stu Jun 12 '16 at 10:01
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This is similar to Can humans control rotation of the Earth? and How can you find the impact necessary to change the direction of Earth's spin?

Suppose the ball is kicked against the direction in which the Earth is rotating. This increases the speed of rotation of the Earth, reducing the rotation period (ie the length of 1 day).

The angular momentum imparted to the Earth by the ball's kick is :
$\Delta L$ = mass of ball x its velocity x radius of Earth = $mvR$.
This is related to the moment of inertia J of the Earth and the increase in angular speed $\Delta \omega$ by
$\Delta L = J\Delta \omega$.
Assuming the Earth is a sphere of uniform density with mass M and radius R then
$J = \frac25MR^2$.
The angular speed $\omega$ is related to the period of rotation of the Earth $T$ by $\omega=\frac{2\pi}{T}$ so $\Delta \omega = - \frac{2\pi}{T^2} \Delta T$.

Bringing this all together and rearranging we get
$mvR = (\frac25MR^2)(-\frac{2\pi}{T^2} \Delta T)$
$\Delta T = - \frac{5mvT^2}{\pi MR}$.
The LHS is the decrease (- sign) in the time it takes for the Earth to make one rotation.

You will need to look up values for M and R, and to convert $T=24$ to hours to seconds. While $vT \approx 2000km$, the distance travelled by the ball during one revolution of the Earth, is comparable with $R \approx 6000km$, the radius of the Earth, there is an enormous difference in the two masses $m$ and $M$. So $\Delta T$ will be very small compared with $T$.

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  • $\begingroup$ Really appreciate the answer. Does the T on the left not cancel out the T on the right? $\endgroup$ – Dr Stu Jun 12 '16 at 10:57
  • $\begingroup$ No, it doesn't. If you take $T$ from LHS to RHS you get $T^2$ on RHS. The reason I put $\frac{\Delta T}{T}$ on the left is because it is often easier to appreciate the significance of a fraction than an absolute amount. However, $\Delta T$ in seconds will be small enough anyway and easy to compare mentally with T=24 hours. (I have edited. Hope this removes any confusion.) $\endgroup$ – sammy gerbil Jun 12 '16 at 11:05
  • $\begingroup$ Understood, many thanks, I'll get the calculator out and see what I get. $\endgroup$ – Dr Stu Jun 12 '16 at 12:50
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First off I would recommend a very large boot. "Size billion" comes to mind.

Second I would query "in what direction are you planning to kick said ball with said billion sized boot, fine sir?" Since direction is not specified in this...ahem...matter...ahem...I find the issue at hand confounding indeed.

Still...we all could use a few more hours in our day here at Lake Wobegon so I think I speak for all parties when I say you'd goal is indeed a laudable one.

While I am not at liberty at this time to offer membership in our secret Lunar Lasoe Society I wish to say your incite into this matter has drawn you to the attention of our avuncular Committee of Desiree's and look forward to more pronouncements.

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  • $\begingroup$ Thank you for your most entertaining answer. I would give your answer an upvote but, alas, I have yet to earn the necessary 'reputation' to have my feelings acknowledged. Accept this as a smiley face in lieu of my up-voting inability. $\endgroup$ – Dr Stu Jun 12 '16 at 9:58

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