How Do You Calculate the Mass and Radius of Planet Physics?

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To calculate the mass and radius of Planet Physics, the discussion revolves around the equations of motion and gravitational forces. The initial calculations for the radius were incorrect due to confusion between the acceleration at the surface and at the altitude of the orbiting ship. The correct approach involves using gravitational equations, where the surface gravity is calculated as 8.8 m/s², leading to a radius of approximately 5.31E6 m. However, discrepancies arise as the textbook states the radius should be 1.33E6 m, prompting further verification of the problem's parameters. The conversation highlights the importance of correctly applying gravitational principles to solve for planetary characteristics.
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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?


Homework Equations





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?
 
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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.
 
ohhhh... of course. Thanks
 
bcjochim07 said:
ohhhh... of course. Thanks
A pleasure :smile:
 
wait a second, If I divide my answer by 2, I still don't get the right answer, which is 1.33E6 m
 
bcjochim07 said:
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.
 
Would the gravity be one fourth of what it is at the surface? Nope... that doesn't work either.
 
I really am not sure what to do
 
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
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
bcjochim07 said:
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.

bcjochim07 said:
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.
 
  • #12
That's why I multiplied my expression for R by 2
 
  • #13
bcjochim07 said:
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.
 
  • #14
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
Is this entered online? I get that as the answer for the radius.
 
  • #16
no, it's the answer in the back of my textbook
 
  • #17
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
no, It's from the back of my textbook
 
  • #19
Kurdt said:
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.
 
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  • #20
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.
 
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