Mass of a planet?

1. Jan 21, 2004

Kacper

mass of a planet???

I need some real good, and quick help. How do you find the mass of a planet with this information? The planet has a satellite 5.2 times 10 to the eighth meters away, and it takes 33 days to circle the planet. Help.

2. Jan 21, 2004

Janus

Staff Emeritus
Just use the formula:

$$T= 2 \pi \sqrt { \frac{r^{3}} {GM}}$$

Which is the formula for finding the period of a satellite, and solve for M.

3. Jan 21, 2004

marcus

I agree with Janus and want to add an extra comment. If you apply this formula to the Earth-Moon system it will give you the combined mass of the earth and the moon.

It does not give the mass of the primary.

But if you are allowed to assume that the mass of the satellite is negligible then the combined mass is essentially the same as the mass of the primary. In that sense it gives what you are looking for.

The mass of the Earth's moon is 1/81 of the mass of Earth if I remember right.
So if you use this formula and get the combined mass you still have
to shave off something like 1/82 of the total if you want the mass of the earth.

4. Jan 21, 2004

Kacper

What do you mean by combined mass?
And if it is the combined mass of both of them I still do not know the ratio. Like you said about the earth being 1/81.

5. Jan 21, 2004

Kacper

I followed these steps and came out with 1.43x10^13. Does anyone agree or did I do something wrong? This is the mass of the planet.

Weighing the Sun
To keep object moving in circle requires a force towards center.
Force is proportional to object's speed squared: the faster its
moving the greater force required. (Escape Velocity)
For planets, the force that keeps them in orbit is gravitational
attraction with sun.
Force of gravity is proportional to Sun's mass

Thus, planet's speed is proportional to Sun's mass
So, Measure planets speed while orbiting to find Sun's mass
One way to measure planet's speed is to measure its orbital
period and orbital size.
speed = orbit circumfrance/ Period
So 1/period is related to mass of sun.

Last edited: Jan 21, 2004
6. Jan 21, 2004

marcus

Dont panic. The teacher probably said that the satellite was small.
Or he meant to say that the satellite mass was negligible compared with the primary.

That formula gives the sum of the two masses (the primary plus the satellite)
but it is customary to NEGLECT the mass of the satellite, and assume it is small and does not matter.
So then the formula give us (to good approximation) the mass of the primary.

In the case of the earth moon system, the earth is 81/82 of the combined mass
and the moon is 1/82 of the combined mass
and the earth mass is 81-times bigger.
But who cares?
To a first approximation 81/82 is the same as 1
so just use the formula like Janus says.

It is a great formula and the man who found it was born
in Weil-der-Stadt in Schwaben, not too far from Stuttgart where you live. Use it.

7. Jan 21, 2004

marcus

33 days is 2.85 million seconds

Janus (or maybe it was Johannes, which sounds similar) says

$$M = \frac{4\pi^2 R^3}{G T^2}$$

G is 6.674E-11 in SI units
r is 5.2E8 meters

$$M = \frac{4\pi^2 5.2E8^3}{6.674E-11* 2.85E6^2}$$

I did not check your number 1.43E13 kilograms. Why should it be wrong?
Oh no!!! I just did check and I got a different number.
Maybe I am making an arithmetic mistake (or you could be?)

remember to put in the unit, if you are working in SI units.

Last edited: Jan 21, 2004
8. Jan 21, 2004

Kacper

ok I am going to try the first equatiom you gave me. I am still confused (only a little) about why you ignore the satellite mass no matter the size. Also didn't Newton make a proportion problem that is something like: F is prop to Mass 1 x Mass 2 over distance.

9. Jan 21, 2004

NateTG

M is the sum of the masses. The trick is that the mass of the satelite is tiny compared to the mass of the Planet, so it's too small to mater.

10. Jan 21, 2004

marcus

Be reasonable and make a rough estimate. This 33 days is rougly similar to our moon's period of some 29+ days

and this distance 520,000 kilometers is roughly similar to the distance to our moon

So shouldnt the mass of the Teacher's Planet be similar to the mass of earth?

But the mass of earth is roughly 1025 kilograms

(more exactly Earth mass is 6E24 kilo, but I mean rough order-of-magnitude)

how can you get 1013? has something terrible happened to the german school system (the way it did already to the USA school system )?

Try again

11. Jan 21, 2004

Kacper

I am horrible at physics so thanks for being patient. I solved the problem using you equation and I got 2.92x10^31. Is that in the ballpark or am I terribly of? Also I am a military dependent living overseas since my dad is station in Germany, so I am in a corrupted American school.

12. Jan 21, 2004

marcus

You mentioned weighing the sun. To weigh the sun you can use the same formula. Make up your mind whether to do it by proportions (which can be tricky) or by brute force SI metric arithemtic. In SI metric it is a no-brainer.

A year is 3.15E7 seconds.
R = 150 million km = 1.5E11 meters.

$$M = \frac{4\pi^2 R^3}{G T^2}$$

G is 6.674E-11 in SI units

$$M = \frac{4\pi^2 1.5E11^3}{6.674E-11* 3.15E7^2} kg$$

Last edited: Jan 21, 2004
13. Jan 21, 2004

marcus

In that case you have suffered enough indignities already.

The handbook mass of the sun is about 2E30 kilogram
that is, in case you do not like the "E" notation,
2 x 1030 kilogram

I believe the planet mass is right around 1025 kilogram.

this is in the same ballpark as the earth which is 6 x 1024

please see if, by putting the T and r into the formula
you get something like the mass I just mentioned

do you have a calculator which handles exponential notation
or socalled "scientific notation"? that makes all this sort of
automatic

14. Jan 21, 2004

Kacper

I got that mass of the sun thing from a book and it said that it worked with any orbiting mass around another larger mass. I resolved the problem and got 1.025x10^25 kg. I do not care about proportions I was just wondering because that is how my friend solved it, but if it is more confusing then I want to solve this problem the way you are teaching me right now before I try anything else. I do have a question about an equation though. I just found this one in my notes can it work? M=(Velocity squared)times distance all divided by Big G.
Big G is the 6.67x10^-11.

15. Jan 21, 2004

Kacper

I have a graphing calculator (ti-83)

16. Jan 21, 2004

marcus

that's great
it is better than my Casio, which has gotten so low
on battery lately I sometimes cannot see the screen

so you CAN do operations like (6.674E-11)* (3.16E7)2

17. Jan 21, 2004

marcus

Yes ideed!
that is a real good one to use
there is hope yet, in spite of all the damage sustained by the schools

18. Jan 21, 2004

Kacper

ok but how do I find the velocity?

19. Jan 21, 2004

marcus

Try it on the sun.

You know already the sun mass is 2E30 kilo

I tell you, or you can figure out, that the earth speed V is 3E4 meters per second

And R is 1.5E11 meters

So sun mass is

$$M = \frac{V^2 R}{G}= \frac{3E4^2 1.5E11}{6.674E-11}= ?$$

It should work out.

It is applicable to circular orbits, so there is a steady speed V, and it gives you the combined mass. But in this case the mass of the earth is negligible. So the sun's mass is essentially the same as the combined mass. The usual rigamarole about "negligible" "essentially" "approximate". But hey it works.

Last edited: Jan 21, 2004
20. Jan 21, 2004

Kacper

Oh ok I am most definite that I got it now. Thank you very much for your time and help. I need to logoff because my Dad needs the computer (its 9:00pm here). Thank you.