What would the tension in the cable be?

  • Thread starter Thread starter fusionxtc
  • Start date Start date
  • Tags Tags
    Cable Tension
Click For Summary

Homework Help Overview

The discussion revolves around a physics problem involving the tension in a hypothetical massless cable that holds the moon in its orbit. Participants are given parameters such as the moon's orbital period, distance from Earth, and its mass, but express concerns about missing information necessary to solve for tension.

Discussion Character

  • Mixed

Approaches and Questions Raised

  • Participants discuss calculating the velocity of the moon and using it in the tension equation. There are questions about unit conversions, particularly converting the orbital period from days to seconds. Some participants express confusion about the necessary formulas and the overall approach to the problem.

Discussion Status

There is ongoing exploration of the problem, with some participants attempting to clarify their understanding of the concepts involved. Guidance has been offered regarding unit conversions and the relationship between tension and centripetal force, but no consensus has been reached on how to proceed effectively.

Contextual Notes

Participants note a lack of familiarity with physics concepts and express feelings of being overwhelmed by the complexity of the problem. There is an acknowledgment of the need for foundational knowledge in physics to tackle such questions.

fusionxtc
Messages
5
Reaction score
0
1. Suppose the moon were held in its orbit not by gravity but by the tension in a massless cable. You are given that the period of the moon's orbit is T = 27.3 days, the mean distance from the Earth to the moon is R = 3.85 x 108 m, and the mass of the moon is M = 7.35 x 1022 kg. What would the tension in the cable be?



2. how do i find the tension when there seems to be missing info. (to me)



3. First i tried finding velocity to put that into the tension equation. v = 3.85 x 10^8 / 27.3. What i used for an equation for tension. T = 7.35 x 10^22 (1.41 x 10^7 )^2 all over 3.85 x 10^8


please help, i have no clue what to do.
 
Physics news on Phys.org
Welcome to PF.
v = 3.85 x 10^8 / 27.3.

What is that? Radius in m / days?

Surely you know what the circumference of a circle is.
 
fusionxtc said:
1. Suppose the moon were held in its orbit not by gravity but by the tension in a massless cable. You are given that the period of the moon's orbit is T = 27.3 days, the mean distance from the Earth to the moon is R = 3.85 x 108 m, and the mass of the moon is M = 7.35 x 1022 kg. What would the tension in the cable be?
2. how do i find the tension when there seems to be missing info. (to me)
3. First i tried finding velocity to put that into the tension equation. v = 3.85 x 10^8 / 27.3. What i used for an equation for tension. T = 7.35 x 10^22 (1.41 x 10^7 )^2 all over 3.85 x 10^8please help, i have no clue what to do.

Let's agree that the tension will be equal to the centripetal force (the force needed to keep the moon in orbit). The centripetal force is given by

[tex]F = m \frac{v^2}{r}[/tex]

What's going to make you think a little bit is your units. Note that your distances are in meters and the mass in kilograms. You'll need to convert from days to something else (what do you think is appropriate? ) in order to get to Newtons.

Another pertinent question:

How far around is the moon's orbit?

Those comments should get you started...
 
Last edited:
ive never had physics before, and no one has every helped me with it. I don't know anything about physics
 
moons orbit is 27.3 days. the problem doesn't give much info. I am completely lost when it comes to this stuff.
 
fusionxtc said:
moons orbit is 27.3 days. the problem doesn't give much info. I am completely lost when it comes to this stuff.

Well, You're going to want to express the final answer in Newtons. Those a SI units, meaning they use meters, kilograms and seconds. So to start with you will need to convert 27.3 days into so many seconds. That will tell you how long it takes the moon to go once around. Then you have to figure out how far the moon travels in that time (once around). Since it's assumed to be a circular orbit (to make the problem easier). That should be easy (what's the formula for the circumference of a circle?). Then you can compute the velocity of the moon and after that it's "plug and chug" since I gave you the equation.

By the way if the only thing you study is stuff you've studied before, you'll never learn anything new!
 
well thanks for the input, but i still do not understand it... obviously i shouldn't be in college.
there's soo many different formulas for just one question. i don't know where to begin. :\
 
fusionxtc said:
well thanks for the input, but i still do not understand it... obviously i shouldn't be in college.
there's soo many different formulas for just one question. i don't know where to begin. :\

Physics is something that requires time and hard work --much hard work to understand. Most people your age do not realize that confusion is the first step towards learning.
 

Similar threads

Replies
13
Views
2K
  • · Replies 19 ·
Replies
19
Views
3K
  • · Replies 7 ·
Replies
7
Views
3K
  • · Replies 6 ·
Replies
6
Views
3K
  • · Replies 3 ·
Replies
3
Views
2K
  • · Replies 5 ·
Replies
5
Views
2K
Replies
6
Views
3K
  • · Replies 19 ·
Replies
19
Views
4K
  • · Replies 10 ·
Replies
10
Views
6K
Replies
4
Views
5K