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

  • Thread starter Thread starter bcjochim07
  • Start date Start date
  • Tags Tags
    Gravity
Click For Summary

Homework Help Overview

The problem involves calculating the mass and radius of a fictional Planet Physics based on the motion of a rock thrown upwards and the orbital characteristics of a cruise ship. The scenario includes parameters such as the initial velocity of the rock, the time of flight, and the orbital period of the ship.

Discussion Character

  • Exploratory, Assumption checking, Problem interpretation

Approaches and Questions Raised

  • Participants discuss the relationship between gravitational acceleration at the surface and at the altitude of the orbiting ship, questioning the validity of their calculations and assumptions regarding the radius.

Discussion Status

Participants are actively exploring different approaches to derive the radius and mass, with some providing calculations while others express confusion about the relationships between the variables involved. There is no explicit consensus on the correct method or answer yet.

Contextual Notes

Some participants note discrepancies between their results and the answers provided in the textbook, leading to questions about the problem's setup and the accuracy of their calculations.

bcjochim07
Messages
366
Reaction score
0

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?
 
Physics news on Phys.org
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 [itex]2r[/itex].
 
  • #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.
 
Last edited:
  • #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.
 
Last edited:

Similar threads

  • · Replies 8 ·
Replies
8
Views
3K
  • · Replies 4 ·
Replies
4
Views
2K
Replies
3
Views
2K
  • · Replies 1 ·
Replies
1
Views
10K
  • · Replies 2 ·
Replies
2
Views
4K
  • · Replies 7 ·
Replies
7
Views
2K
  • · Replies 3 ·
Replies
3
Views
1K
  • · Replies 1 ·
Replies
1
Views
3K
  • · Replies 4 ·
Replies
4
Views
5K
  • · Replies 2 ·
Replies
2
Views
1K