# Moon in Orbit

## Homework Statement

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 10^8 m, and the mass of the moon is M = 7.35 x 10^22 kg.

What would the tension in the cable be?

## Homework Equations

Centripetal Acceleration = v^2/R

## The Attempt at a Solution

Tension = Ma
a = v^2/R
v = 2pi/T

First I concerted the 27.3 days to seconds which is T = 2.358*10^6. Which means that v = 2.6638*10^-6. Then I plugged that in for a, so a = 1.843*10^-20. So then the Tension should be 7.35*10^22 * 1.843*10^-20 which is 1354.67N. But it is saying that is the wrong answer. What did I do wrong?

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hage567
Homework Helper
Take another look at your equation for velocity. It isn't right.

thanks.. i figured it out :)

i calculated the angular velocity and got w = 2.66*10^-6 and then multiplied that by R to get the acceleration.

hage567
Homework Helper
No, I mean it doesn't make sense. Velocity is m/s right? Where is the "m" in your equation?! If the moon is moving in a circle, what distance does it go in one revolution?

Velocity is the "directional displacement" covered over a period of time. Considering that the direction in this case is perpendicular to the radius vector at all points, you'd have to look a bit closer at v = 2*pi/T...is that really the "directional displacement" (distance in this case) over time?...you're missing a variable.

tony873004