# Homework Help: Two Kepler Questions

1. Jun 17, 2011

### mcdowellmg

Question 1:
1. The problem statement, all variables and given/known data

https://www.physicsforums.com/showthread.php?t=44239 except the planet is Saturn.

2. Relevant equations

$$r = (t / (2pi)) ^ {2 / 3} * (G * M) ^ {1 / 3}$$

3. The attempt at a solution

I tried many things, so I thought I would give tony873004's solution attempt from the thread I linked to a shot.

"where r is the radius of your orbit (distance from the center of Saturn), t is the period of your orbit, which is the same as the rotational period of Saturn (38,745 seconds), G is the gravitational constant $$6.67 * 10^{-11}$$, M is the mass of Saturn ($$5.6851 * 10^{26} kg$$).

Subtract from this the radius of Saturn (60,268 km) to get your altitude."

I calculated from all of that 456,708 km, which is apparently wrong.
Am I doing something wrong here with finding the right rotational period or mass of Saturn?

The second question is:

1. The problem statement, all variables and given/known data

A 25.0-kg satellite has a circular orbit with a period of 2.35 h and a radius of 7.30×10^6 m around a planet of unknown mass. If the magnitude of the gravitational acceleration on the surface of the planet is 9.10 m/s2, what is the radius of the planet?

3. The attempt at a solution

GMm/r^2 = m*v^2/r
v = 2*pi*r/T
so GM/r^2 = (4*pi^2*r)/T^2

So the mass of the planet M = 4*pi^2*r^3/(G*T^2) = (4*pi^2*(7.30*10^6)^3)/(6.67x10^-11*(2.35*3600)^2)= 3.22*10^24 kg

b) Now Using the law of gravitation we get GMm/R^2 =m*g at the surface we get

GM/R^2 = g So R = sqrt(GM/g) = sqrt(6.67x10^-11*3.22*10^24/9.10) = 4.89*10^6

That is somehow also wrong. Are the formulas set-up correctly?

2. Jun 17, 2011

### Staff: Mentor

For Question 1 your method looks okay, but your answer is not correct. Probably a calculation slip-up. Try running the numbers again. If you don't make headway, post some partial results that can be checked.

For Question 2, is your result to be specified in meters or in kilometers?

3. Jun 17, 2011

### mcdowellmg

The second question is to be answered in meters.

I tried the first question again, and got 112,732 km, which is much different! Is that more in line with what you were calculating? It is difficult dealing with all of this scientific notation.

Here is how I calculated. $$r = (t / (2pi)) ^ {2 / 3} = (38,745 seconds / 6.2832...) = 6166.44... ^ {2/3} = 336.27$$
$$(G * M) ^ {1 / 3} = (6.67 * 10 ^ {-11} (m ^ {3})/(kgs ^ {2} * 5.6851 * 10^{26} kg = (37919617000000000) ^ {1/3} = 335960.32$$
$$336.27 * 335960.32 = 1.13*10 ^ {8}$$, which is in meters, so it would be 113,000 km.
$$113,000 km - 60,268 km (the radius of Saturn) = 52,732 km.$$

Well, now I have gotten 52,732 km! I swear that I have the worst ability at using calculators/Wolfram Alpha. I need to go back to doing everything manually, if only I could remember how!

Last edited: Jun 17, 2011
4. Jun 17, 2011

### Staff: Mentor

I obtained a smaller value (very roughly half) for the first question.

For the second question, the value I obtained using the same numbers that you did differed in the second decimal place. So, maybe run through the numbers again.

5. Jun 17, 2011

### mcdowellmg

I might as well try the calculation again for question #2 on here, step-by-step.

$$(G*M*m)/(r ^ {2}) = (m*v ^ {2})/r$$
$$v = (2*\pi*r)/T$$
Therefore, $$(G*M)/r ^ {2} = (4*\pi ^ {2}*r)/T ^ {2}$$
So, the mass of the planet, $$M=(4*\pi ^ {2}*(7.30*10 ^ {6}) ^ {3})/(6.67*10 ^ {-11}*(2.35 years * 3600 seconds) ^ {2}) = (1.54*10 ^ {22})/0.00477382572 = 3.23*10 ^ {24}$$

Now Using the law of gravitation we get $$G*M*m/Radius of planet ^ {2} =m*g$$ at the surface we get

$$GM/R^2 = g$$ So, $$R = \sqrt{GM/g}= \sqrt{(6.67*10 ^ {-11}*3.23*10 ^ {24})/9.10)} = 333931335560000000/9.10 = 4.87*10 ^ {6}$$ meters

Is that right?

Last edited: Jun 17, 2011
6. Jun 17, 2011

### Staff: Mentor

The numbers look good this time.