1
$\begingroup$

Let us suppose that Earth and Jupiter are orbiting around the Sun in same orbit, If we talk about at any particular point.

(Which is not possible but for the sake of question)

I was wondering that if they were orbiting in same orbit then will they both have same time period? If yes, then why because as they both have different angular momentum and both have so much of differences.

$\endgroup$
4
$\begingroup$

I was wondering that if they were orbiting in same orbit then will they both have same time period? If yes, then why because as they both have different angular momentum and both have so much of differences.

I'll break this down into two parts, first looking at the period of individual objects orbiting the Sun at a distance of one astronomical unit (but ignoring the presence of other objects in the same orbit), and then looking at what would happen if Jupiter happened to be in the same orbit as the Earth.

To first order, it doesn't matter whether you're asking about a grain of sand, a big rock that's a kilometer across, an Earth-sized object, a Jupiter-sized object, or a one solar-mass neutron star. The universality of free-fall says all of these objects will fall sunward with the same acceleration. There's a second order effect that can't be ignored for larger objects: The Sun is also falling toward the object.

A grain of sand and the big rock are both tiny compared to the Sun. They will both have orbits that are very, very close to Keplerian, with the period given by $2\pi\sqrt{r^3/u_\odot}$, where $\mu_\odot$ is the Sun's gravitational parameter. (Aside: Conceptually, $\mu_\odot=GM_\odot$ where $G$ is the universal gravitational constant and $M_\odot$ is one solar mass. In practice, scientists know $\mu_\odot$ to more than ten places of accuracy but they only know $G$ and $M_\odot$ to a lousy four places of accuracy. It's best to use $\mu_\odot$.)

Since the mass of the grain of sand is many orders of magnitude smaller than that of the big rock, this is a good place to answer your secondary question (If yes, then why because as they both have different angular momentum and both have so much of differences). The answer lies in the universality of free-fall. While the gravitational force on the two objects are very different, their accelerations are identical. Since both are so many, many orders of magnitude smaller than the Sun, the acceleration of the Sun toward the grain of sand is essentially zero, and the same applies to the big rock.

Those second order effects become noticeable when the mass of the orbiting object is appreciably larger than the measurement error in $\mu_\odot$. Scientists know $\mu_\odot$ to over ten places of accuracy. The Earth's mass is about 3×10-6 solar masses, so that's not negligible. Jupiter's mass is about 1/1000 solar masses, so that's not negligible at all. That simple formula for orbital period becomes $2\pi\sqrt{r^3/(\mu_\odot+\mu_\text{planet})}$ in the case of an orbiting planet. Plugging in the numbers means that an Earth-sized object located one astronomical unit from the Sun would orbit the Sun 2000 times in 2000 years. A Jupiter-sized object at the same distance would orbit the Sun 2001 times in that same time span. Two thousand years is the blink of an eye on an astronomical time scale.

Finally, I'll address the last part of the question. If Jupiter and the Earth occupied the same orbit, that wouldn't be the case for long. The interaction between Jupiter and the Earth would be very strong every 2000 years or so, the period at which Jupiter overtakes the Earth. In very short order (a small multiple of 2000 years), Jupiter would expel the Earth from the solar system.

$\endgroup$
  • $\begingroup$ I must say the explanation you have is so understandable and easy to build up the concept and to clear our doubts related to this question. +1 :) $\endgroup$ – Shashank Apr 27 '15 at 18:55
3
$\begingroup$

The rotation period $T$ is given by $$T=2\pi \sqrt{\dfrac{a^3}{G(M_\text{Sun}+M_\text{planet})}}$$ where $a$ is the sum of the half axes of the ellipse.

Routhly:

  • $M_\text{Sun}=2\times 10^{30}$ kg
  • $M_\text{Earth}=6\times 10^{24}$ kg
  • $M_\text{Jupiter}=2\times 10^{27}$ kg

If you assume both Earth and Jupiter are orbiting around the Sun (and neglect the forces between Jupiter and the Earth) on the orbit of the Earth, then $a=150\times 10^9$ m.

For both cases, $M_\text{Sun}\gg M_\text{Earth}$ and $\gg M_\text{Jupiter}$ so the difference will be very little. The calculation will give around 365 days.

So the time periods would no be significantly different. Now, if you also consider the force between Jupiter and the Earth, the problem is much more complicated!

$\endgroup$
  • $\begingroup$ What you have written is right ,so can I say that Time period will be same for both ? or just because of slightly different it will not same ... means my point is what's the conclusion? $\endgroup$ – Shashank Apr 27 '15 at 15:04
  • $\begingroup$ You have a difference of 5%. That is up to you to neglect it! If you are looking for a qualitative behaviour you can say it is similar, if you want to calculate a trajectory for your satellite (in our imaginary world where Jupiter and the Earth share the same orbit) you'd better consider the difference. It's like asking me if we can say that 0.25 dollar is the same as 0 dollar. It depends on the context! Mathematically the are for sure different. For a millionaire physicist, the are probably the same. For a poor physicist the are probably not the same. $\endgroup$ – anderstood Apr 27 '15 at 15:09
  • 2
    $\begingroup$ Your figure for the mass of Jupiter is wrong $\endgroup$ – John Rennie Apr 27 '15 at 15:34
  • $\begingroup$ @ShashankSharma Notice the correction of the value of $M_\text{Jupiter}$. You can now say that there is no significant change of value of the period. $\endgroup$ – anderstood Apr 27 '15 at 15:46
  • 1
    $\begingroup$ @DavidHammen: good answer though. I must admit my initial reaction was just "yes". We get used to considering the Sun as infinitely massive. $\endgroup$ – John Rennie Apr 27 '15 at 16:31

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.