# Calculating Tension in the Cable to Hold the Moon

• Naeem
The conversation is about calculating the tension in a hypothetical situation where the moon is held in its orbit by a massless cable instead of gravity. We are given the period of the moon's orbit, its distance from Earth, and its mass. To find the tension in the cable, we use the formula T=Ma, where M is the mass of the moon and a is the acceleration. To find a, we use the formula a=v^2/R, where v is the velocity and R is the radius. We can find v by converting the period of 27.3 days to seconds and using the formula v=2π/T. In summary, to find the tension in the cable, we need to use the mass, period, and
Naeem
Q.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.
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What would the tension in the cable be?

'T' in Newtons.

My idea:

T = Ma
mass of the moon is given to us.

a = v^2 /R

To find v , we can v = 2 pi /T

Radius is also given to us.

We can convert the 27.3 days to seconds, by 23.7 * 24 hours * 3600 seconds.

Well is my approach correct.

Hi Naeem,

this looks purfect to me.

Your approach is correct. To find the tension in the cable, we need to use the centripetal force equation:

F = ma = mv^2/R

Where F is the centripetal force, m is the mass of the moon, v is the velocity of the moon, and R is the distance between the moon and the Earth. Since we are assuming the cable is massless, the tension in the cable is equal to the centripetal force.

To find v, we can use the formula v = 2πR/T, where T is the period of the moon's orbit. Plugging in the given values, we get:

v = 2π(3.85 x 10^8 m)/(27.3 days * 24 hours * 3600 seconds) = 1.02 x 10^3 m/s

Now, we can plug in the values for m, v, and R into the centripetal force equation:

F = (7.35 x 10^22 kg)(1.02 x 10^3 m/s)^2/(3.85 x 10^8 m) = 1.97 x 10^20 N

Therefore, the tension in the cable to hold the moon in its orbit would be approximately 1.97 x 10^20 Newtons. This is an enormous amount of force, which shows how strong the force of gravity is in keeping the moon in its orbit around the Earth.

## 1. How is the tension in the cable to hold the moon calculated?

The tension in the cable to hold the moon is calculated using the formula: T = (m x v^2) / r, where T is the tension, m is the mass of the moon, v is the velocity of the moon, and r is the radius of the moon's orbit.

## 2. What is the mass of the moon used in the calculation?

The mass of the moon used in the calculation is approximately 7.35 x 10^22 kilograms. This is the commonly accepted mass of the moon in scientific calculations.

## 3. Is the velocity of the moon a constant value in the calculation?

No, the velocity of the moon is not a constant value in the calculation. It varies depending on the moon's position in its orbit. The average velocity is calculated by dividing the circumference of the moon's orbit by its orbital period.

## 4. How is the radius of the moon's orbit determined?

The radius of the moon's orbit is determined by measuring the distance between the center of the moon and the center of the Earth. This distance is approximately 384,400 kilometers.

## 5. Are there any other factors that may affect the tension in the cable?

Yes, there are other factors that may affect the tension in the cable, such as the gravitational pull of other celestial bodies in the solar system and the Earth's rotation. However, these factors are relatively small compared to the mass and velocity of the moon and can be considered negligible in the calculation.

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