# Keplerian Orbits

1. May 24, 2006

### Mirielle

Hi, I have a question on calculating the period of a planet where I know the gravity at the equator and at the pole.

Here is the known information for the planet:
Planet radius: $6502 km$
Planet mass: $4.71 \ast 10^{24} kg$

An explorer, who weighs 191 lbs on earth, stands at the equator of this mystery planet. He finds that he currently weighs 159 lbs.
He then moves to the north pole of the planet and finds that he weighs 105% of what he previously weighed.
The question then asks what the period of rotation for the planet is.

So, I thought the following formula would probably be useful in solving this.
$$v = \frac{2 \pi R}{T}$$
Or in this case:
$$T = \frac{2 \pi R}{v}$$
So, I already know the radius, which only leaves figuring out the velocity with which it is spinning...

$$a_{equator}= \frac{Gm}{r^2}$$

$$a_{pole}= 1.05 a_{equator}$$

We also know that:
$$a_{centrip.}= \frac{v^2}{r}$$

So, now I solve for v by combining these...

$$\frac{1.05 GM }{r^2 }= \frac{v^2 }{r }$$

which gives me:

$$v = \sqrt{\frac{1.05 GM}{r}}$$

Now I plug in the following values from above:

$$r = 6502 km = 6502000 m$$

$$m = 4.71 \ast 10^{24} kg$$

(And the constant)

$$G = 6.673 \ast 10^{-11}$$

And I find that v = 7125 m/s

Now, I convert this to hours due to the period being in hours.
v = 7125 m/s = 25650000 m / hr

Then we return to this equation I presented before:

$$T = \frac{2 \pi R}{v}$$

And plugging in the values...

$$T = \frac{2 \pi (6502000)}{25650000} = 1.59 hr$$

However, this is wrong.
I'm curious to know where I'm going wrong here.

P.S. this is the first time I've ever used TeX, I hope it works properly!

Last edited: May 24, 2006
2. May 24, 2006

### Andrew Mason

Suggestion: Set out the equation for T in algebraic form and then plug in numbers.

In this case, the difference between the weight at the equator and at the pole is the centripetal force due to the rotation of the planet. So you don't really have to worry about what the force of gravity is.

$$m\Delta g = F_c = m\omega^2r = m\frac{4\pi^2}{T^2r}$$

We know what his weight is on earth and g for the earth, so we can determine his mass. From that you can determine the value for g on the equator and at the pole

$$\Delta g = \frac{4\pi^2}{T^2r}$$

$$1/T = \sqrt{r\Delta g/4\pi^2}$$

AM

Last edited: May 24, 2006
3. May 24, 2006

### vsage

The weight at that point is the sum of two forces. You tried to use centripetal force to explain the entire weight. I'll leave it up to you to fill in the rest.

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