Question

A man is standing on an international space station which is orbiting Earth at an altitude 520km with a constant speed $7.6\,km/s$. If the men's weight is $50\,kg$, his acceleration is
A. $7.6\,km/{s^2}$
B. $7.6\,m/{s^2}$
C. $8.4\,m/{s^2}$
D. $10\,m/{s^2}$

Answer 1

Hint: Using the universal law of gravitation and the force equation from Newton’s second law, we can find the expression for finding acceleration due to gravity. By comparing the equation at the surface of the earth and at the given altitude, we can get the value of acceleration due to gravity experienced by the man.

Complete step by step answer:
The speed of the space station in which the man is standing is given as \[7 \cdot 6\,km/s\]
Altitude of space station is given as $520\,km$
Weight of the man is
 $w = 50\,kg$
We need to find the acceleration due to gravity.
Here the only force acting on the space station is the gravitational force of the earth.
So, the centripetal force required for the motion of the space station in its orbit is provided by this gravitational force.
We know that the gravitational force is given as
$F = \dfrac{{GmM}}{{{r^2}}}$ ……………………..(1)
Where $m$ is the mass of the man, M is the mass of earth r is the radius of the orbit.
We know the force is the product of mass and acceleration. That is
$F = ma$
Let the acceleration due to gravity at this distance be $g'$.
Then we can write force as
$F = mg'$ ……………………….(2)
On equating equation (1) and equation (2), we get,
$ \Rightarrow mg' = \dfrac{{GmM}}{{{r^2}}}$
$ \Rightarrow g' = \dfrac{{GM}}{{{r^2}}}$
Similarly, on the surface of the earth, we have $g = \dfrac{{GM}}{{{R^2}}}$ where $R$ is the radius of the earth.
On dividing $g$ with $g'$, we get
$ \Rightarrow \dfrac{g}{{g'}} = \dfrac{{\dfrac{{GM}}{{{R^2}}}}}{{\dfrac{{GM}}{{{r^2}}}}} = \dfrac{{{r^2}}}{{{R^2}}}$
$ \Rightarrow g' = g{\left( {\dfrac{R}{r}} \right)^2}$
The radius of the orbit can be found by adding the radius of earth with the altitude of the space station. That is,
$r = R + 520$
$ \Rightarrow r = 6400 + 520 = 6920\,km$
On substituting the given values. we get,
$ \Rightarrow g' = 9 \cdot 8 \times {\left( {\dfrac{{6400 \times {{10}^3}}}{{6920 \times {{10}^3}}}} \right)^2}$
On simplification of the above terms, we get
$ \Rightarrow g' = 8 \cdot 38\,m/{s^2}$
$\therefore g' \approx 8 \cdot 4\,m/{s^2}$
This is the acceleration due to gravity at the space station.

Therefore, the correct answer is option (C).

Note:
We know that the gravitational force is inversely proportional to the square of the distance between the objects. Hence the value of acceleration due to gravity will decrease as we move higher and higher above the earth’s surface. So, the value of acceleration experienced by the man in the space station, $g'$ will be less than the value of gravity on the earth’s surface.