What would the tension in the cable be?

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

The discussion focuses on calculating the tension in a hypothetical massless cable holding the moon in orbit, using the moon's orbital period of T = 27.3 days, a mean distance of R = 3.85 x 108 m, and the moon's mass of M = 7.35 x 1022 kg. Participants emphasize the need to convert the orbital period from days to seconds to find the velocity and subsequently apply the centripetal force formula, F = m(v2/r), to determine the tension. The importance of using SI units (meters, kilograms, seconds) is highlighted for accurate calculations.

PREREQUISITES
  • Understanding of centripetal force and its formula
  • Knowledge of unit conversions, particularly from days to seconds
  • Familiarity with the circumference of a circle
  • Basic physics concepts related to orbital mechanics
NEXT STEPS
  • Learn how to convert time units from days to seconds for physics calculations
  • Study the formula for the circumference of a circle to calculate orbital distance
  • Explore the concept of centripetal force in more depth, including examples
  • Practice solving problems involving gravitational forces and orbital mechanics
USEFUL FOR

Students studying physics, particularly those interested in orbital mechanics, as well as educators and tutors looking for examples of tension and centripetal force calculations.

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

F = m \frac{v^2}{r}

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
2K
  • · Replies 7 ·
Replies
7
Views
2K
  • · 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
5K
Replies
4
Views
5K