Orbit time change?

Classical Mechanics Level 3

This problem is about Newtonian physics and involves imaginary (as in just made up) situations.

The object being orbited is perfectly spherical. It is uniformly dense throughout. The orbit is perfectly circular at the very surface of the object in all cases. The test object and the object being orbited are the entire universe for this problem. The test object is effective a mathematical point in the sense that it has no effect on the position of the orbited object. Perfect vacuum conditions exist to the very surface of the orbited object. You may assume Issac Newton's shell theorem without proof.

I suggest a review of the physics involved.

The orbit time is $2\pi\sqrt{\frac Rg}$ . Using metric values, the Earth's mean radius is 6.3710088 Mm and the acceleration due to gravity is 9.80665 $\frac{m}{s^2}$ . The result is 1 hour, 24 minutes, 24.35 seconds. (That is about as many digits as justified by the input parameters.).

For this problem, the distance unit is 1 R for the standard sized object's radius, and the surface orbital period around that object is 1 T time unit. The uniform density is such that this happens. If it helps, then use the Earth as your model, 1 R is 1 Earth mean radius and 1 T is 1 hour, 24 minutes and 24.35 seconds.

The question: How does the surface orbital period change as an orbited object's radius changes from one half R to 2 R (note definition of R above) ?" Note: the assumptions, the density and the gravitational constant do not change. Only the orbited object's radius changes.

Surface orbit times remain the same. Not enough information has been given to answer the question. Surface orbit times increases. Surface orbit times decreases.

1 solution

A Former Brilliant Member
Feb 26, 2019

This problem can be done using a dimensional analysis alone. If the radius of the orbited object changes by a factor of $f$ , then, in the fraction $\frac Rg$ , the numerator changes by $f$ and, this is the important part, so does the denominator, as volume of the orbited object changes by $f^3$ and the changed radius changes the force of gravity per mass unit by $f^{-2}$ for a combined factor of $f$ ; therefore, $\frac Rg$ does not change. Q.E.D.

Orbit time change?

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