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If the value of the gravitational constant $G$ starts to decrease, which of the following will happen?

a) The length of the year will increase.

b) The Earth will follow a spiral path of decreasing radius.

c) The kinetic energy of the Earth will decrease.

This was an unsolved question from my textbook. The answer given is option a and c.

I could not free the variables like the orbital period and velocity from each other so to reach a point from where I could proceed. I am a class 12$^{\rm th}$ student and would like to get a reasonable explanation using Newtonian mechanics.

Where I got stuck:

The orbital velocity of Earth is given by $V=\sqrt{GM/R}$. So, the velocity and radius are dependent on each other.

Also, the orbital period is $T=2\pi\sqrt{R^3/GM}$.

Again, the same problem. What to do?

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    $\begingroup$ Could you provide your attempt at a solution? That would probably show us how to answer the question in a way useful to you ... $\endgroup$ – Sanya Aug 8 '16 at 15:00
  • $\begingroup$ Well,what I tried was equating the gravitational force to the centripetal force. But, this did not do the job. Then I wondered if the energy would remain conserved, and could find no reason that it should be. I stuck at this point. If that confuses you, you may try using keplers law to arrive directly where I was stuck. $\endgroup$ – Amritansh Singhal Aug 8 '16 at 15:08
  • $\begingroup$ @AmritanshSinghal Maybe it is a good starting point to think about what happens if we set $G$ to zero. What would happen then to the kinetic energy of the earth and its trajectory? $\endgroup$ – Jannick Aug 8 '16 at 15:12
  • $\begingroup$ If $G$ started decreasing, I'd be more concerned about dying than what Earth's orbit is doing $\endgroup$ – Jim Aug 8 '16 at 15:27
  • $\begingroup$ @jim. C'mon... but you'll die then anyway. Better if you try figure out what would be happening $\endgroup$ – Amritansh Singhal Aug 8 '16 at 15:29
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The Earth is in a steady, (nearly) circular orbit. To maintain this orbit, a radial acceleration of $a = v^2/{\rm R}_\oplus$ is required. The gravitational acceleration of the Earth is $F/m = G{\rm M}_\odot/{\rm R}_\oplus^2$. Putting these two equations together gives:

$$v = \sqrt{\frac{G{\rm M}_\odot}{{\rm R}_\oplus}}$$

Seems like you got this far in your attempt. Now what? Perhaps the simplest way to proceed is a bit of more qualitative reasoning. You can think of $G$ as a constant that defines the strength of the gravitational force. If $G$ starts to decrease, the force holding Earth on its circular orbit starts to decrease, and it no longer has the necessary radial acceleration to hold it at its current radius. In other words, it is moving too fast for a circular orbit at its current radius; it must move out (i.e. ${\rm R}_\oplus$ increases). The motion of Earth now has a radial component (outward), and the only force is acting radially (inward), so unless $G$ has instantaneously become $0$, Earth must be slowing down. Kinetic energy is simply $\frac{1}{2}{\rm M}_\oplus v^2$, so answers (a) and (c) are true.

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