# Homework Help: Stars and gravitation

1. Apr 2, 2005

### jaymode

Here is my problem:
The star is 59 light years from the earth and has a mass of 1.90×1030 kg.

A. A large planet of mass 1.20×1028 kg is known to orbit this star. The planet is attracted to the star by a force of 3.50×1026 N when their centers are separated by a distance equal to the semi-major axis of the planet's orbit. Calculate this distance (in m).

B. What is the period (in days) of this planet's orbit?

C.Other as yet undiscovered planets may orbit this star. If a second planet is in a circular orbit around the star with an orbital period equal to 8.00 times the value found in part (b), what is the radius (in m) of the second planet's orbit?

I have attempted A, but cannot seem to figure out how to get the distance. I was trying to use the equation a = (GmT^2/4pi^2)^1/3

2. Apr 2, 2005

### SpaceTiger

Staff Emeritus
Try just Newton's Law on part A:

$$F=\frac{GMm}{r^2}$$

Then use your equation for part B.

3. Apr 2, 2005

### jaymode

That does not work out. I get an incredibly small number which I know is wrong.

4. Apr 2, 2005

### jaymode

nm I was stupid and not doing the math correctly.

5. Apr 2, 2005

### whozum

For a)

$$F = \frac{GMm}{R^2}$$

$$R = \sqrt{\frac{GMm}{F}}$$

$$R = \sqrt{\frac{6.67x10^{-11} * 1.9 x 10^{30} * 1.2 x 10^{28}}{3.5 x 10^{26}}} = \sqrt{4.35 x 10^{21}} = 6.59 x 10^{11} m$$

No
For b)

a = (GmT^2/4pi^2)^1/3

$$a = \frac{3.5x10^{26}}{1.2x10^{28}}$$

G = 6.67 x 10^-11
T = What you want

$$\sqrt{\frac{4a^3\pi^2}{Gm}} = T$$

For c)

Find T from above, multiply by 8. Solve for R and find R.