What would the tension in the cable be?

In summary: It's important that you continue to ask questions and seek clarification from more experienced individuals, in order to continue growing as a student. In summary, the tension in a massless cable holding the moon in its orbit would be equal to the centripetal force. The centripetal force is given by F = m \frac{v^2}{r} where m is the mass of the moon and r is the radius of the orbit. The moon travels around the Earth in a circular orbit and it takes 27.3 days for it to complete one orbit. The circumference of a circle is circumference (in meters) multiplied by the number of days in a year (365). The moon's velocity is also easy to find, as v
  • #1
fusionxtc
5
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
  • #2
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.
 
  • #3
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:
  • #4
ive never had physics before, and no one has every helped me with it. I don't know anything about physics
 
  • #5
moons orbit is 27.3 days. the problem doesn't give much info. I am completely lost when it comes to this stuff.
 
  • #6
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!
 
  • #7
well thanks for the input, but i still do not understand it... obviously i shouldn't be in college.
theres soo many different formulas for just one question. i don't know where to begin. :\
 
  • #8
fusionxtc said:
well thanks for the input, but i still do not understand it... obviously i shouldn't be in college.
theres 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.
 

1. What is tension in a cable?

Tension in a cable is the force that is applied to the cable in order to keep it taut. It is the pulling force that is exerted on both ends of the cable.

2. How is tension in a cable calculated?

Tension in a cable is calculated by dividing the force applied to the cable by the cross-sectional area of the cable. This can be represented by the equation T = F/A, where T is the tension, F is the force, and A is the cross-sectional area.

3. What factors affect the tension in a cable?

The tension in a cable is affected by the amount of force applied, the length and weight of the cable, and the angle at which the cable is being pulled. Other factors such as temperature and material properties can also impact tension.

4. Why is tension important in cables?

Tension is important in cables because it helps to distribute the weight and forces applied to the cable evenly. It also helps to keep the cable stable and prevents it from sagging or breaking.

5. How can tension in a cable be adjusted?

Tension in a cable can be adjusted by changing the force applied to the cable, adjusting the angle at which the cable is being pulled, or changing the material or diameter of the cable. It is important to carefully calculate and adjust tension to ensure the cable can safely support the weight and forces applied to it.

Similar threads

  • Introductory Physics Homework Help
Replies
19
Views
785
  • Introductory Physics Homework Help
Replies
7
Views
2K
  • Introductory Physics Homework Help
Replies
3
Views
1K
  • Introductory Physics Homework Help
Replies
5
Views
2K
  • Introductory Physics Homework Help
Replies
10
Views
4K
  • Introductory Physics Homework Help
Replies
6
Views
3K
  • Introductory Physics Homework Help
Replies
19
Views
3K
  • Introductory Physics Homework Help
Replies
4
Views
4K
  • Introductory Physics Homework Help
Replies
6
Views
3K
  • Introductory Physics Homework Help
Replies
9
Views
2K
Back
Top