bcjochim07

## Homework Statement

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?

## 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?

Staff Emeritus
Gold Member
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.

bcjochim07
ohhhh... of course. Thanks

Staff Emeritus
Gold Member
ohhhh... of course. Thanks
A pleasure

bcjochim07
wait a second, If I divide my answer by 2, I still don't get the right answer, which is 1.33E6 m

Staff Emeritus
Gold Member
wait a second, If I divide my answer by 2, I still don't get the right answer, which is 1.33E6 m
The acceleration that you calculated earlier is the acceleration on the surface, i.e. at r, whereas your cruise ship is orbiting at 2r.

bcjochim07
Would the gravity be one fourth of what it is at the surface? Nope... that doesn't work either.

bcjochim07
I really am not sure what to do

Staff Emeritus
Gold Member
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.

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.

Staff Emeritus
Gold Member
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]

This part is good.

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.

The ship is still at $2r$.

bcjochim07
That's why I multiplied my expression for R by 2

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

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

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

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

bcjochim07
no, it's the answer in the back of my textbook

Staff Emeritus
Gold Member
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?

bcjochim07
no, It's from the back of my textbook

Homework Helper
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?

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:
Homework Helper
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: