Name: Introduction to Laws of Planetary Motion
Uploaded: 2019-07-19
Duration: 8 min 11 s
Description: Did you know that planets move around the sun in elliptical orbits? On this elliptical path, planets move faster at one portion and slower at another. Find o...

Question

A spacecraft is circling the earth along a circular orbit and retains its orientation with respect to the earth. Is zero gravity inside the spacecraft absolute in this case?

Accepted Answer

The fact that the spacecraft retains its orientation with respect to the earth means that all points of the spacecraft move with the same angular velocity. Suppose that the point closest to the surface of the earth moves with the orbital (satellite) velocity $(ω)$ according to the equation

$m ω^{2} R = G \frac{M m}{R^{2}}$

$\Rightarrow ω^{2} R = G \frac{M}{R^{2}} . . . (1)$

where $R$ is the distance between this point and the centre of the earth.

The point of the spacecraft farthest from the earth moves with an acceleration $ω^{2} (R + D),$ where $D$ is the distance between the two points.

If we consider the spacecraft to be a non-inertial system, we can assume that on an object of mass $m$ placed at the point farthest from the earth there acts a force of inertia(Known as centrifugal force.)

$F_{1} = - m ω^{2} (R + D) .$

At the same time, there is the force of gravity acting on this object:

$F = G \frac{M m}{(R + D)^{2}}$

'The sum of these two forces plays the role of "weight" for the object, or numerically the reaction of the support exerted on the object:

$F_{w} = m ω^{2} (R + D) - G \frac{M m}{{(R + D)}^{2}}$

Bearing in mind that $D ≪ R,$ We can replace $(R + D)^{- 2}$ with $\frac{(\frac{1 - 2 D}{R})}{R^{2} .}$ Thus

$F_{w} \approx m [ω^{2} R (1 + \frac{D}{R}) - G \frac{M}{R^{2}} (1 - 2 \frac{D}{R})]$

$\Rightarrow F_{w} \approx m [ω^{2} R (\frac{D}{R}) + G \frac{M}{R^{2}} (2 \frac{D}{R})]$

And if we allow for $(1),$ we get

$F_{w} \approx G \frac{M m}{R^{2}} (\frac{3 D}{R})$

Since $\frac{G M m}{R^{2}}$ is equal, to a high accuracy, to the weight of the object on the surface of the earth, or $m g,$

we get $F_{w} = \frac{3 D}{R} m g$

This expression gives the "weight" of an object in the spacecraft at the point farthest from the earth. Assuming that $D$ is $2.1 m$ and bearing in mind that $R = 6300 km,$ we find that the "weight" of an astronaut whose mass is $70 kg$ is $6.9 \times 10^{- 4} N$ at the point within the spacecraft farthest from the earth.

L.A.Sena Solutions for Chapter: Fundamentals of Mechanics, Exercise 1: Exercise

L.A.Sena Physics Solutions for Exercise - L.A.Sena Solutions for Chapter: Fundamentals of Mechanics, Exercise 1: Exercise

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