Another gravity problem Please help

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In summary, the conversation is about a problem where a person is trying to calculate the mass and radius of a planet based on information about the acceleration on its surface and the orbit of a cruise ship around it. The conversation includes various calculations and attempts at solving the problem, but the correct answer is not obtained.
  • #1
bcjochim07
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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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  • #2
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
ohhhh... of course. Thanks
 
  • #4
bcjochim07 said:
ohhhh... of course. Thanks
A pleasure :smile:
 
  • #5
wait a second, If I divide my answer by 2, I still don't get the right answer, which is 1.33E6 m
 
  • #6
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.
 
  • #7
Would the gravity be one fourth of what it is at the surface? Nope... that doesn't work either.
 
  • #8
I really am not sure what to do
 
  • #9
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:

1. What is the difference between mass and weight in terms of gravity?

Mass is the amount of matter an object contains, while weight is the measure of the force of gravity acting on an object. Mass remains constant regardless of location, while weight changes depending on the strength of gravity in that location.

2. How does the gravitational force between two objects change as their distance increases?

The gravitational force between two objects is inversely proportional to the square of the distance between them. This means that as the distance between the objects increases, the gravitational force decreases.

3. What is the formula for calculating the force of gravity between two objects?

The formula for calculating the force of gravity between two objects is F = G (m1m2)/r^2, where G is the gravitational constant, m1 and m2 are the masses of the two objects, and r is the distance between them.

4. How does the mass of an object affect its gravitational pull?

The greater the mass of an object, the greater its gravitational pull. This means that larger objects have a stronger gravitational pull than smaller objects.

5. How does the strength of gravity differ on different planets?

The strength of gravity on a planet depends on its mass and size. The larger and more massive a planet is, the stronger its gravitational pull. This means that gravity is stronger on larger planets like Jupiter, and weaker on smaller planets like Mercury.

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