
A satellite is in elliptic orbit around the earth with apogee of $6R$ and perigee of $2R$ where $R = 6400\,Km$ radius of the earth. Find the velocity of the satellite at apogee in $Km{s^{ - 1}}.$ $[G = 6.67 \times {10^{ - 11}}\,N{m^2}K{g^{ - 2}}{\kern 1pt} {\kern 1pt} {\kern 1pt} ,{\kern 1pt} {\kern 1pt} M = 6 \times {10^{24}}Kg]$
Answer
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Hint: In order to solve this question, we should know that, a satellite revolves around the earth either in circular orbit or elliptical orbit under the centripetal force acting on it which is balanced by gravitational force of attraction between satellite and earth, we will use the formula of velocity of satellite revolving around earth at given distance from the earth.
Formula used:
If $G$ is the gravitational constant, $M$ is the mass of earth and $r$ is the distance between earth and the satellite revolving around it in elliptical orbit by considering earth it’s one of the foci then velocity at that point is calculated as
$v = \sqrt {\dfrac{{GM}}{r}} $
Complete step by step answer:
According to the question we have given that,
Mass of the earth, $M = 6 \times {10^{24}}Kg$
Gravitational constant, $G = 6.67 \times {10^{ - 11}}\,N{m^2}K{g^{ - 2}}$
Converting the value of $G$ in Kilometres units by putting $1{m^2} = {10^{ - 6}}K{m^2}$ we get,
$G = 6.67 \times {10^{ - 17}}\,NK{m^2}K{g^{ - 2}}$
The distance where velocity is to be measured, $r = 6R$
Radius of the earth, $R = 6400\,Km$
Using the formula of velocity of satellite as,
$v = \sqrt {\dfrac{{GM}}{r}} $
Putting values of all parameters we get,
$v = \sqrt {\dfrac{{6.67 \times {{10}^{ - 17}} \times 6 \times {{10}^{24}}}}{{6 \times 6400}}} $
On solving we get,
$v = \sqrt {\dfrac{{6.67 \times {{10}^5}}}{{64}}} $
$\therefore v = 3.23\,Km{s^{ - 1}}$
Hence, the velocity of the satellite at apogee is $v = 3.23\,Km{s^{ - 1}}$.
Note: It should be remembered that, while solving such questions always convert all physical quantity values in same SI units and basic unit of conversion used as $1\,m = 0.001\,Km$ also, apogee is the point where satellite is at farthest point from earth whereas perigee is the point where satellite is very close to the earth.
Formula used:
If $G$ is the gravitational constant, $M$ is the mass of earth and $r$ is the distance between earth and the satellite revolving around it in elliptical orbit by considering earth it’s one of the foci then velocity at that point is calculated as
$v = \sqrt {\dfrac{{GM}}{r}} $
Complete step by step answer:
According to the question we have given that,
Mass of the earth, $M = 6 \times {10^{24}}Kg$
Gravitational constant, $G = 6.67 \times {10^{ - 11}}\,N{m^2}K{g^{ - 2}}$
Converting the value of $G$ in Kilometres units by putting $1{m^2} = {10^{ - 6}}K{m^2}$ we get,
$G = 6.67 \times {10^{ - 17}}\,NK{m^2}K{g^{ - 2}}$
The distance where velocity is to be measured, $r = 6R$
Radius of the earth, $R = 6400\,Km$
Using the formula of velocity of satellite as,
$v = \sqrt {\dfrac{{GM}}{r}} $
Putting values of all parameters we get,
$v = \sqrt {\dfrac{{6.67 \times {{10}^{ - 17}} \times 6 \times {{10}^{24}}}}{{6 \times 6400}}} $
On solving we get,
$v = \sqrt {\dfrac{{6.67 \times {{10}^5}}}{{64}}} $
$\therefore v = 3.23\,Km{s^{ - 1}}$
Hence, the velocity of the satellite at apogee is $v = 3.23\,Km{s^{ - 1}}$.
Note: It should be remembered that, while solving such questions always convert all physical quantity values in same SI units and basic unit of conversion used as $1\,m = 0.001\,Km$ also, apogee is the point where satellite is at farthest point from earth whereas perigee is the point where satellite is very close to the earth.
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