Question

A body is suspended from a spring balance kept in a satellite. The reading of the balance is $W_1$ when the satellite goes in a an orbit of radius $R$ and is $W_2$ when it goes in an orbit or radius $2R$.

• A. $W_1=W_2$
  
• B. $W_1<W_2$
  
• C. $W_1>W_2$
  
• D. $W_1\ne W_2$
  

Answer 1

The correct option is A $W_1=W_2$



The orbital velocity of a satellite revolving around a planet of mass $M$ in an orbit of radius $r$ is given by $$v=\sqrt{\frac{GM}{r}}$$Since, the velocity of the body inside the satellite will be same as the orbital velocity of the planet, so the centrifugal force acting on the body will be,
$F_c=\frac{m{v}^2}{r}$

$\Rightarrow F_c=\frac{m{\left(\sqrt{\frac{GM}{r}}\right)}^2}{r}$

$\Rightarrow F_c=\frac{GMm}{r^2}...\left(1\right)$

Gravitational force acting on the body will be,

$F_g=\frac{GMm}{r^2}...\left(2\right)$

So, from free body diagram for a body hanging in the satellite,

$T+F_c=F_g$ ​

where, $T$ is the tension produced in spring balance.

Substituting the values from equation $\left(1\right)$ and $\left(2\right)$,

$\Rightarrow T+\frac{GMm}{r^2}=\frac{GMm}{r^2}$

$\Rightarrow T=0$

If weight $W_1$ is measured by the spring balance, then

As, for spring balance $T=W_1=0$

Thus, tension is zero in all cases.

Hence ,$W_1=W_2=0$.

So, option (a) is correct answer.

$\text{Alternate Solution:}$
For a satellite revolving around the earth at any orbital radius, the gravitational force towards earth is countered by centrifugal force $F_g=F_c$

So, for a body hanging in the satellite,

$W_1+F_c=F_g$

$\Rightarrow W_1=0$

Similarly, $W_2=0.$

Thus, $W_1=W_2$ ​ .

Hence, option (a) is the correct answer.

Why this question: Any satellite in its circular orbit will not apply any reaction force and hence weightlessness will be felt inside a satellite.