Question

The critical speed of a satellite of mass $500\text{ kg}$is $\text{20 }{\text{m}}/{\text{s}}\;$. What is the critical speed of a satellite of mass $1000\text{ kg}$moving in the same orbit?
(A). $\text{10 m}{{\text{s}}^{-1}}$
(B). $\text{20 km h}{{\text{r}}^{-1}}$
(C). $\text{72 m}{{\text{s}}^{-1}}$
(D). $\text{72km h}{{\text{r}}^{-1}}$

Answer 1

Critical velocity is the speed with which a satellite rotates in it’s orbit. Using the formula for critical velocity, ${{\text{V}}_{\text{c}}}$ solve for $500\text{ kg}$mass satellite. Substitute the missing values and compare for $1000\text{ kg}$mass. Convert the units of speeds to check the options and get the right answer.

Formula used:
${{\text{V}}_{\text{c}}}\text{ = }\sqrt{\dfrac{\text{GM}}{\text{R}}}$

Complete step-by-step answer:
The critical speed is the horizontal speed given to a satellite so that it can be put into a stable circular orbit around the earth. It is also called orbital velocity. It is denoted by ${{\text{V}}_{\text{c}}}$ .
The formula for ${{\text{V}}_{\text{c}}}$is-
${{\text{V}}_{\text{c}}}\text{ = }\sqrt{\dfrac{\text{GM}}{\text{R}}}$ - (1)
Where $\text{G}$= gravitational constant, $6.67\times {{10}^{-11}}$ $\text{N}{{\text{m}}^{\text{2}}}\text{k}{{\text{g}}^{\text{-2}}}$
$\text{M}$= mass of Earth, $\text{6 }\\!\\!\times\\!\\!\text{ 1}{{\text{0}}^{\text{24}}}\text{kg}$
$\text{R}$ = radius of Earth, $\text{6 }\\!\\!\times\\!\\!\text{ 1}{{\text{0}}^{\text{3}}}\text{km}$
It is given to us that ${{\text{V}}_{\text{c}}}$= $\text{20 m}{{\text{s}}^{-1}}$ for a satellite of mass $500\text{ kg}$
From eq (1) we can see that ${{\text{V}}_{\text{c}}}$ does not depend on the mass of satellite, which means that it will remain constant irrespective of the mass of satellite. So, the value of ${{\text{V}}_{\text{c}}}$ for a satellite of mass $1000\text{ kg}$ is $\text{20 }{\text{m}}/{\text{s}}\;$.
Converting the units of ${{\text{V}}_{\text{c}}}$from $\dfrac{\text{m}}{\text{s}}\text{ to }\dfrac{\text{km}}{\text{hr}}$, we get,
$\text{20 }{\text{m}}/{\text{s}}\;\text{ = }\dfrac{\dfrac{20}{1000}\text{km}}{\dfrac{1}{3600}\text{hr}}$
$\Rightarrow \text{ 20 m}{{\text{s}}^{-1}}\text{ = 72 km h}{{\text{r}}^{-1}}$
Therefore, the correct option is (D). $\text{72 km h}{{\text{r}}^{-1}}$.

So, the correct answer is “Option D”.

Additional Information: The centripetal force for the circular motion of satellites around the earth is provided by the gravitational force acting on it due to the Earth.

Note: The height at which satellite orbits above the surface of the earth is ignored as it is negligible in comparison to the radius of the Earth. Other formulas for ${{\text{V}}_{\text{c}}}$ are $\sqrt{\text{gR}}$ (where $\text{g}$is acceleration due to gravity), $\text{2R}\sqrt{\dfrac{\text{G }\\!\\!\pi\\!\\!\text{ }\\!\\!\rho\\!\\!\text{ }}{\text{3}}}$ (where $\text{ }\\!\\!\rho\\!\\!\text{ }$ is the density of Earth). From the given formulas we can conclude that orbital velocity remains constant near the surface of earth.