Classical Mechanics Challenge

[Martin]

Imagine a planet that is so small and unmassive that, even close in, the orbit speed is only 6.7 miles per hour

(if there were someone running, they could keep up with you, or more precisely the runner would go into orbit as well)

 

perhaps it has a specially benign atmosphere which offers no air resistance but it good to breathe and is a comfortable shirtsleeve sort of temperature

 

maybe it is spring and you can smell plumblossom and magnolia and you are in orbit just grazing the hilltops, seeing everything on the planet at the speed of a run.

 

You are invited to calculate the mass of this planet, just from one additional piece of information. In an orbit with a steady speed of 6.7 mph it takes

one and 7/8 hours to go full circle around the planet.

Calculate the mass any units you please. I've stated it in common units so it should make no difference.

[Martin]

what I am looking for is an elegant and efficient calculation.

for an experienced person (perhaps that includes anybody who might be interested) writing the answer out in symbols would be trivial or nearly so. but

what I want is an actual calculation showing what the mass of the planet is and how you found it.

[Janus]

Since nobody else is biting:

 

The most elegant solution is to use the formula:

 

[math]M = \frac{v^3t}{G}[/math]

Where v is the orbital velocity and t is the orbital period.

 

Converting to SI units this gives:

[math]\frac{(2.98)^3 (6750)}{2 \pi 6.673x10^{-11}\frac} = 4.26x10^{14} kg[/math]

 

The above formula is easily derived from the the fact that if we know the orbital velocity and we know the period, we can get the circumference of the orbit from

 

[math]C = vt[/math]

 

We can then get the radius by dividing by 2 pi:

 

[math]r =\frac{vt}{2 \pi}[/math]

 

This radius by the way turns out to be about 3.2 km.

 

The equation for orbital velocity is

 

[math]v = \sqrt{\frac{GM}{r}}[/math]

 

we simply substitute for r:

 

[math]v =\sqrt{\frac{2 \pi GM}{vt}}[/math]

 

and rearrange to solve for M.

 

We can now also solve for the density of the body. We know its mass and we can calculate its volume from its radius, giving us a volume of 1.37 x 10^11 m³.

 

Dividing the mass by the volume we get the density of 3100 kg/m³. This is a density of a little less than that of our moon.

[Martin]

Since nobody else is biting:

 

The most elegant solution is to use the formula:

 

[math]M = \frac{v^3t}{G}[/math]

Where v is the orbital velocity and t is the orbital period.

 

Converting to SI units this gives:

[math]\frac{(2.98)^3 (6750)}{2 \pi 6.673x10^{-11}\frac} = 4.26x10^{14} kg[/math]

 

...

 

This radius by the way turns out to be about 3.2 km.

 

...

Dividing the mass by the volume we get the density of 3100 kg/m³. This is a density of a little less than that of our moon.

 

Janus' date=' thanks for responding. I think your approach is elegant and efficient, I differ only in one minor detail: I think for your formula you meant to write had a 2pi in denominator.

 

"[math']M = \frac{v^3t}{2\pi G}[/math]

Where v is the orbital velocity and t is the orbital period."

 

and indeed that is what you used for the calculation, where I see the 2pi.

 

the reason I appreciate your responding is that I want to try out some natural units and make a side-by-side comparison solving the same problem in metric (as you did) and in natural units.

 

Natural units have some convenience and some awkward aspects. Doing some parallel calculation is just the thing to bring these out.

 

I got the same answer: I could write it 0.43 x 10¹⁵ kg, to show the agreement with your answer. But I got it in different terms (a variant of the Planck mass unit). I have to go out now but will get back later this evening and show the same problem worked in natural units terms.

[Martin]

You are invited to calculate the mass of this planet, just from one additional piece of information. In an orbit with a steady speed of 6.7 mph it takes

one and 7/8 hours to go full circle around the planet.

Calculate the mass any units you please. I've stated it in common units so it should make no difference.

 

in the system of natural units I am trying out, to a fair approximation:

6.7mph = E-8

and the orbit period P = 112.5 minutes = 0.25E47

 

using the same equation as you did, the arithmetic comes to

 

M = 4 P v³ = 4 x 0.25E47 x E-24 = E47 x E-24 = E23

 

to give a human scale interpretation of this mass we can use a "pound" mass of 0.434 kg = E8, and then the planet mass is

E15 "pounds". Or approximately 0.43 x 10¹⁵ kilograms, in agreement with your answer!

= 4.26x10^{14} kg[/math]

...

This radius by the way turns out to be about 3.2 km.

...

Dividing the mass by the volume we get the density of 3100 kg/m³. This is a density of a little less than that of our moon.

 

 

I also confirm your figure for the radius---I got around 2 miles----and with your estimate of the density---about same as earth's moon.