Calculating Tension in the Cable to Hold the Moon

[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.

 

[quasar987]

Hi Naeem,

this looks purfect to me.

 

[thepatientmental]

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.