Question

An artificial satellite revolves around the earth at a height of $1000{\text{ km}}$. The radius of earth is $6.38 \times \;\,{10^3}\;{\text{km}}$. Mass of the earth $= 6 \times {10^{24}}{\text{kg}}$; $G = 6.67 \times {10^{11}}{\text{ N}}{{\text{m}}^2}{\text{k}}{{\text{g}}^{{\text{ - 2}}}}$. Find the orbital speed and period of revolution of the satellite.
(A) $7364{\text{ m}}{{\text{s}}^{ - 1}},6297{\text{s}}$
(B) $1064{\text{ m}}{{\text{s}}^{ - 1}},6297{\text{s}}$
(C) $9364{\text{ m}}{{\text{s}}^{ - 1}},9297{\text{s}}$
(D) $8364{\text{ m}}{{\text{s}}^{ - 1}},7297{\text{s}}$

Answer 1

When a satellite is thrown into the orbit of the earth, it experiences gravitational pull. And when the centrifugal force is balanced out by the gravitational force, it starts to revolve around the earth in a certain orbit.
The velocity with which a satellite revolves around the earth is known as orbital velocity. And Time taken to complete one revolution around the earth by satellite is known as the Time period.

Complete step by step answer:
Write the expression of the orbital speed ${v_0}$ for an artificial satellite,
${v_0} = \sqrt {\dfrac{{GM}}{{R + h}}}$
Here, $M$ is the mass of earth, $G$ is the gravitational constant, $R$ is the radius of the earth and $h$ is the height of the satellite from the earth.
Substitute $6.38 \times \;\,{10^3}\;{\text{km}}$ for$R$, $6.67 \times {10^{11}}{\text{ N}}{{\text{m}}^2}{\text{k}}{{\text{g}}^{{\text{ - 2}}}}$ for$G$, $6 \times {10^{24}}{\text{kg}}$ for $M$ and $1000{\text{ km}}$ for$h$.

{v_0} = \sqrt {\dfrac{{(6.67 \times {{10}^{11}}{\text{ N}}{{\text{m}}^2}{\text{k}}{{\text{g}}^{{\text{ - 2}}}})(6 \times {{10}^{24}}{\text{kg)}}}}{{(6.38 \times \;\,{{10}^3}\;{\text{km}}\left( {\dfrac{{{{10}^3}{\text{m}}}}{{1{\text{km}}}}} \right) + 1000{\text{ km}}\left( {\dfrac{{{{10}^3}{\text{m}}}}{{1{\text{km}}}}} \right){\text{)}}}}} \\\ = 7364\;{\text{m}}{{\text{s}}^{ - 1}} \\\ \ $$ Write the expression for the Period of revolution $$T$$ of satellite. $$T = \dfrac{{2\pi (R + h)}}{{{v_0}}}$$ Substitute $$6.38 \times \;\,{10^3}\;{\text{km}}$$ for$$R$$,$$1000{\text{ km}}$$ for $$h$$ and $$7364\;{\text{m}}{{\text{s}}^{ - 1}}$$ for $${v_0}$$ $$\ T = \dfrac{{2\pi (6.38 \times \;\,{{10}^3}\;{\text{km}}\left( {\dfrac{{{{10}^3}{\text{m}}}}{{1{\text{km}}}}} \right) + 1000{\text{ km}}\left( {\dfrac{{{{10}^3}{\text{m}}}}{{1{\text{km}}}}} \right))}}{{7364\;{\text{m}}{{\text{s}}^{ - 1}}}} \\\ = 6297{\text{ s}} \\\ \ $$ Therefore,$$7364\;{\text{m}}{{\text{s}}^{ - 1}}$$, $$6297{\text{ s}}$$are orbital speed and period of revolution of the satellite respectively. **So, the correct answer is “Option A”.** **Note:** The expression for the orbital velocity is used to calculate the orbital velocity of the satellite and the expression for the Time period is used to calculate the period of revolution. Standardize the units of the values used in the expressions of Orbital velocity and time period.