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

Hint: 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.