1. Apr 18, 2008

### bcjochim07

1. The problem statement, all variables and given/known data
While visiting Planet Physics, you toss a rock straight up into the air at 11m/s and catch it 2.5 s later. While you visit the surface, your cruise ship orbits at an altitude equal to the radius of the planet every 230 min. What are the mass & radius of Planet Physics?

2. Relevant equations

3. The attempt at a solution

0m=0m + (11m/s)(2.5s) + .5a(2.5s)^2
ac=-8.8

230 min = 13800s angular velocity = 2pi rad/ 13800 = 4.55E-4 rad/s

ac= angular velocity^2 * r
8.8= (4.55E-4)* r
r=4.25 * 10 ^7 m This is not the correct answer. Could someone please explain what I am doing wrong?

2. Apr 18, 2008

### Hootenanny

Staff Emeritus
Note that your altitude is equal to the radius of the planet, which means that your cruise ship is 2r from the centre of the planet.

3. Apr 18, 2008

### bcjochim07

ohhhh.... of course. Thanks

4. Apr 18, 2008

### Hootenanny

Staff Emeritus
A pleasure

5. Apr 18, 2008

### bcjochim07

wait a second, If I divide my answer by 2, I still don't get the right answer, which is 1.33E6 m

6. Apr 18, 2008

### Hootenanny

Staff Emeritus
The acceleration that you calculated earlier is the acceleration on the surface, i.e. at r, whereas your cruise ship is orbiting at 2r.

7. Apr 18, 2008

### bcjochim07

Would the gravity be one fourth of what it is at the surface? Nope... that doesn't work either.

8. Apr 18, 2008

### bcjochim07

I really am not sure what to do

9. Apr 18, 2008

### Kurdt

Staff Emeritus
You're equating two things that are incompatible. The acceleration at the height of the ship is not the same as at the surface of the planet. If you consider setting up two equations in terms of the mass of the planet and the radius, then you can eliminate a variable.

10. Apr 18, 2008

### bcjochim07

Ok, here's what I tried:

since at the surface of the planet
g= GM/R^2
R= sqrt[ GM/g]

I calculated g at the surface to be 8.8 m/s^s

so I substituted that into the equation:
R= sqrt[GM/8.8]

Then I move on to the orbit which is 2R from the center

So the g of the ship would be

gship= GM/(2*sqrt[GM/8.8])^2

I get gship= 2.2 m/s^2

But using the equation a= angular velocity^2 * r
I again get a wrong value for r.

11. Apr 18, 2008

### Kurdt

Staff Emeritus
This part is good.

The ship is still at $2r$.

12. Apr 18, 2008

### bcjochim07

That's why I multiplied my expression for R by 2

13. Apr 18, 2008

### Kurdt

Staff Emeritus
It was that bit to which I was referring in my previous post. Sorry for not making that clear.

14. Apr 18, 2008

### bcjochim07

2.2 m/s^2 = (4.55E-4)^2*r
r= 1.06E7
this r is the distance between the ship and the center of the planet

So then I divide by 2 to get the radius of the planet: 5.31E6 m

But that's not the correct answer

15. Apr 18, 2008

### Kurdt

Staff Emeritus
Is this entered online? I get that as the answer for the radius.

16. Apr 18, 2008

### bcjochim07

no, it's the answer in the back of my textbook

17. Apr 18, 2008

### Kurdt

Staff Emeritus
Are you sure the question is as it is stated the book? Failing that I'll get someone else to check that I haven't missed something by mistake. What is the answer in the book out of curiosity?

18. Apr 18, 2008

### bcjochim07

no, It's from the back of my textbook

19. Apr 19, 2008

### Shooting Star

I get the radius as ~ 5.31E6 m.

EDIT: Sorry, bcjochim07. I didn't see that you've already given the answer in post #5.

Last edited: Apr 19, 2008
20. Apr 19, 2008

### alphysicist

Hi Shooting Star,

In post #5 he said the book's answer was 1.33e6 m. I got the same answer that you and he did.

Last edited: Apr 19, 2008