Question

A satellite is moving in a circular orbit at a certain height above the earth's surface. It takes 5.26$\times$${10^3}$s to complete one revolution with a centripetal acceleration equal to 9.32 $m{s^{ - 2}}$. The height of satellite orbiting above the earth is
(Earth's radius =6.37$\times$${10^6}$m)
A) 220 km
B) 160 km
C) 70 km
D) 120 km

Answer 1

Hint:In an easy way to solve the question is first write what is given, find out the equation for velocity from the centripetal formula, equate it with the time period of revolution and take out the height. Do not forget that the total radius will be the radius of earth and the height of the satellite.

Complete step by step answer:
Step 1:
Centripetal acceleration: is the property of the motion of a body traversing a circular path. The acceleration is directed radially toward the centre of the circle and has a magnitude equal to the square of the body's speed along the curve divided by the distance from the centre of the circle to the moving body.
We are given:
A satellite is moving in a circular orbit at a certain height above the earth's surface.
It takes T= 5.26$ \times {10^3}$s to complete one revolution.
The centripetal acceleration equal to 9.32 $m{s^{ - 2}}$ = a
We need to find the height of the satellite orbiting above the earth.
Step 2:
Now, coming to the solution part:
The centripetal acceleration is given by $a = \dfrac{{{v^2}}}{r}$ here, r is the radius of earth and height of satellite = $\left( {{R_e} + h} \right)$
Putting the value of a and rearranging for v we get,$v = \sqrt {9.3\left( r \right)} $
We can rewrite the equation as $v = \sqrt {9.3\left( {{R_e} + h} \right)} $, from here we will find h.
Time period of revolution is given by $T = \dfrac{{2\pi \left( {{R_e} + h} \right)}}{v}$
Substituting the value of v we get, $T = \dfrac{{2\pi \left( {{R_e} + h} \right)}}{{\sqrt {9.3\left( {{R_e} + h} \right)} }}$
We are given the time period of one revolution. Substituting the value of T we get,
$ \Rightarrow $ 5.26$\times{10^3}$s=$\dfrac{{2 \times 3.14\sqrt {\;{R_e} + h} }}{{3.05}}$
Simplifying the above equation: $\sqrt {{R_e} + h} = 2.55 \times {10^3}$
This implies, ${R_e} + h$ =$\left( {2.55 \times {{10}^3}} \right)^2$ $ \Rightarrow 6.53 \times {10^6}$
Now radius of earth ${\operatorname{R} _e}$ =$6.37 \times {10^6}m$, then h=$6.53 \times {10^6} - 6.37 \times {10^6}$ m
From there we will get h=160 km
Hence, the height of the satellite orbiting above the earth is 160 km.

Option B is correct.

Note: Point of mistake:
Here in this question, in step 2: there are many values which are written in power form and also some equations are there which are equated with some other, for example the equation of velocity is equated by the time period to find h. So be cautious while calculating as there may occur mistakes while solving for height and for the roots.