A remote-sensing satellite of earth revolves in a circular orbit at a height of \[0.25\times {{10}^{6}}m\] above the surface of the earth. If earth’s radius is \[6.38\times {{10}^{6}}m\] and \[g=9.8m{{s}^{-2}}\] then the orbital speed of the satellite is:
(A). \[6.67km{{s}^{-1}}\]
(B). \[7.76km{{s}^{-1}}\]
(C). \[8.56km{{s}^{-1}}\]
(D). \[9.13km{{s}^{-1}}\]

Hint: For the uniform motion in a circular orbit around the earth, there will be a gravitational force acting on the satellite. This will be equal to the centripetal force acts on the satellite. From these equal forces, we can find out the orbital speed.

Formula used: \[{{F}_{c}}=\dfrac{m{{v}^{2}}}{{{R}_{o}}}\], where m is the mass of the satellite, v is the velocity of the satellite and \[{{R}_{o}}\] is the radius of the orbit in which the satellite is revolving.
\[{{F}_{g}}=\dfrac{GMm}{{{R}_{o}}^{2}}\], where m is the mass of the satellite, M is the mass of the earth, \[{{R}_{o}}\] is the radius of the orbit and G is the gravitational constant.
\[g=\dfrac{GM}{{{R}^{2}}}\], where G is the gravitational constant, M is the mass of the earth and R is the radius of the earth.

Complete step by step answer:

In this question, the remote sensing satellite is revolving around the earth at a height of \[0.25\times {{10}^{6}}m\]. The radius of the earth is \[6.38\times {{10}^{6}}m\].
We can calculate the orbital speed of the satellite from the centripetal force and the gravitational force.
The centripetal force acting on the satellite can be written as,
\[{{F}_{c}}=\dfrac{m{{v}^{2}}}{{{R}_{o}}}\], where m is the mass of the satellite, v is the velocity of the satellite and \[{{R}_{o}}\] is the radius of the orbit in which the satellite is revolving.
The gravitational force acting on the satellite is,
\[{{F}_{g}}=\dfrac{GMm}{{{R}_{o}}^{2}}\], where m is the mass of the satellite, M is the mass of the earth, \[{{R}_{o}}\] is the radius of the orbit and G is the gravitational constant.
According to the satellite motion, these two forces are the same. The centripetal forces acting towards the centre of the circle. So in this case, the gravitational force is the centripetal force.
\[{{F}_{g}}={{F}_{c}}\]
Therefore, we can write as,
\[\dfrac{m{{v}^{2}}}{{{R}_{o}}}=\dfrac{GMm}{{{R}_{o}}^{2}}\]
So, the orbital speed will be,
\[v=\sqrt{\dfrac{GM}{{{R}_{o}}}}\]…………………(1)
Since the satellite is revolving a height above the earth, the radius of the orbit will be,
\[{{R}_{o}}=R+h\], R is the radius of the earth and h is the height of the satellite.
We can substitute this equation into the orbital speed equation.
\[v=\sqrt{\dfrac{GM}{R+h}}\]……………………(2)
As we know, the acceleration due to gravity can be written as,
\[g=\dfrac{GM}{{{R}^{2}}}\], where G is the gravitational constant, M is the mass of the earth and R is the radius of the earth.
\[g{{R}^{2}}=GM\]………………………(3)
We can add this equation into equation (2)
\[v=\sqrt{\dfrac{g{{R}^{2}}}{R+h}}\]
Now we can assign the given information to the equation.
\[v=\sqrt{\dfrac{9.8\times {{(6.38\times {{10}^{6}})}^{2}}}{(6.38\times {{10}^{6}})+0.25\times {{10}^{6}}}}\]
\[v=7756m{{s}^{-1}}\]
\[v=7.76km{{s}^{-1}}\]
So the correct option is B.

Note: Do not treat the orbital radius as the earth radius. We have to add the height of the satellite with the radius of the earth to get the orbital radius. Since the centrifugal force is equal to the gravitational force, we can balance that force instead of equating the centripetal and gravitational forces. This method is also possible