The Hubble Space Telescope (HST) is a scientific instrument of extreme sensitivi

Question

The Hubble Space Telescope (HST) is a scientific instrument of extreme sensitivity. Because of its finite size, the telescope is subject to a tidal force from the planet's gravitational field (much like the way the world's oceans experience tidal forces due to the moon's gravity). If this tidal force is too large, the stress may affect the desired operation of the telescope. In this problem, treat the HST like a uniform cylinder with length L and diameter D orbiting with its center of mass at an altitude h above the surface of Earth, which has radius R and mass M. Find an approximate expression, to first order in the satellite's dimensions (L and/or D), for the ratio of the maximum tidal acceleration of the satellite to the centripetal acceleration of its center of mass.

Explanation / Answer

The tidal forces acting on the Earth from the moon's gravity are due to the fact that the moon is slightly closer to one side of the Earth than the other. The tidal force is the difference between the force acting on the near side and the force acting on the far side of the Earth.

In the same way, you can calculate the difference between the force of gravity acting on the near side and acting on the far side of the satellite. Then take their difference to get the "tidal acceleration". Then take the ratio of that result to the centripetal acc of the satellite.

First, cetrip acc of satellite: we know the force acting on the satellite is GMm / (R+h)^2

so centrip acc = GM / (R+h)^2

Now the tidal acc...

I will define "a" as half the linear dimension (I'll use L... so a = L/2) of the telescope. Then:

Force acting on near side = Gmm / (R+h-a)^2

Force acting on far side = GMm / (R+h+a)^2

difference in force = GMm [ (R+h-a)^-2 - (R+h+a)^-2 ]

tidal acc (drop the "m") = GM [ (R+h-a)^-2 - (R+h+a)^-2 ]

now simplify what is in the brackets. We're going to use the binomial approximation. This says that

if x << 1, then (1+x)^n is approximately 1 + nx

In your case, n is -1, and we're going to factor out R+h from each of the highlighted expressions. So we get

(R+h)^-2 * (1 - x)^-2 - (R+h)^-2 * (1 + x)^-2 where x = a/(R+h)

and this becomes...

(R+h)^-2 [ 1 + 2x - 1 - 2x ] = (R+h)^-2 * [4x]

SO NOW...

tidal acc = GM * 4x / (R+h)^2

And...

ratio of tidal acc to centrip acc =

= [ GM * 4x / (R+h)^2 ] / [ GM / (R+h)^2 ] =

= 4x = 4 a / (R+h) =

= 4 * (L/2) / (R+h) =

= 2 L / (R+h)   

This is a tiny ratio... two times the length of the telescope... divided by the distance from the telescope to the center of the Earth!! It comes out roughly to about 1/200,000 for the HST.

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