Question

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?

Answer 1

Tension in the cable = 200.655 × 10^(18) N

Step-by-step explanation:

The tension in the cable will be the force on the string which is caused by when it holds against the centripetal acceleration.

Now, the formula for centripetal acceleration is given by;

a = v²/R

Now, let's find the velocity.

The circumference is 2πR

We are given that R = 3.85 x 10^(8) m

So, Circumference = 2 × π × 3.85 x 10^(8) = 24.19 × 10^(8) m

We are told the period of the moon orbit is 27.3 days

Distance travelled per day = Circumference/ period

Distance travelled per day = (24.19 × 10^(8))/27.3 = 88608058.608 m/day

Now, a day has 24 hours = 24 × 60 × 60 = 86400 seconds

Thus,

Distance travelled per seconds =

88608058.608/86400 = 1025.556 m/sec

So, from a_c = v²/r,

a_c = 1025.556²/3.85x10^8

a_c = 0.00273 m/s²

Now, Force = Mass x Acceleration = ma

We are given mass = 7.35 x 10^(22) kg

Plugging in the relevant values, we have;

F = 0.00273 x 7.35 x 10^(22)

F = 200.655 × 10^(18) N