Question
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 earth. If earth’s radius is $$6.38 \times {10^6}m$$ and $$g = 9.8\,m{s^{ - 2}},$$ then the orbital speed of the satellite is
A.
$$7.76\,km{s^{ - 1}}$$
B.
$$8.56\,km{s^{ - 1}}$$
C.
$$9.13\,km{s^{ - 1}}$$
D.
$$6.67\,km{s^{ - 1}}$$
Answer :
$$7.76\,km{s^{ - 1}}$$
Solution :
Given, height of a satellite $$h = 0.25 \times {10^6}m$$
Earth’s radius, $${R_e} = 6.38 \times {10^6}m$$
For the satellite revolving around the earth, orbital velocity of the satellite
$$\eqalign{ & {v_0} = \sqrt {\frac{{G{M_e}}}{{{R_e}}}} = \sqrt {\frac{{G{M_e}}}{{{R_e}\left[ {1 + \frac{h}{{{R_e}}}} \right]}}} \cr & \Rightarrow {v_0} = \sqrt {\frac{{g{R_e}}}{{1 + \frac{h}{{{R_e}}}}}} \cr} $$
Substitutes the values of $$g,{R_e}$$ and $$h,$$ we get
$$\eqalign{ & {v_0} = \sqrt {60 \times {{10}^6}} m/s \cr & {v_0} = 7.76 \times {10^3}m/s \cr & = 7.76\,km/s \cr} $$
Given, height of a satellite $$h = 0.25 \times {10^6}m$$
Earth’s radius, $${R_e} = 6.38 \times {10^6}m$$
For the satellite revolving around the earth, orbital velocity of the satellite
$$\eqalign{ & {v_0} = \sqrt {\frac{{G{M_e}}}{{{R_e}}}} = \sqrt {\frac{{G{M_e}}}{{{R_e}\left[ {1 + \frac{h}{{{R_e}}}} \right]}}} \cr & \Rightarrow {v_0} = \sqrt {\frac{{g{R_e}}}{{1 + \frac{h}{{{R_e}}}}}} \cr} $$
Substitutes the values of $$g,{R_e}$$ and $$h,$$ we get
$$\eqalign{ & {v_0} = \sqrt {60 \times {{10}^6}} m/s \cr & {v_0} = 7.76 \times {10^3}m/s \cr & = 7.76\,km/s \cr} $$