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Another gravity problem Please help

  • Thread starter bcjochim07
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1. 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?


2. Homework 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?
 

Answers and Replies

Hootenanny
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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.
 
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ohhhh.... of course. Thanks
 
Hootenanny
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wait a second, If I divide my answer by 2, I still don't get the right answer, which is 1.33E6 m
 
Hootenanny
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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.
 
374
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Would the gravity be one fourth of what it is at the surface? Nope... that doesn't work either.
 
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I really am not sure what to do
 
Kurdt
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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.
 
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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.
 
Kurdt
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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 [itex]2r[/itex].
 
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That's why I multiplied my expression for R by 2
 
Kurdt
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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.
 
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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
 
Kurdt
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Is this entered online? I get that as the answer for the radius.
 
374
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no, it's the answer in the back of my textbook
 
Kurdt
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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?
 
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no, It's from the back of my textbook
 
Shooting Star
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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:
alphysicist
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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:

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