
Two satellites $A$ and $B$ go round the planet $P$ in circular orbits having radii $4R$ and $R$ respectively. If the speed of the satellite $A$ is $3v$,then the speed of satellite $B$ will be
(A) $6v$
(B) $12v$
(C) $\dfrac{{3v}}{2}$
(D) $\dfrac{{4v}}{3}$
Answer
580.8k+ views
Hint:Here first we have to use the relationship between the velocity of a satellite and the radius of the circular orbits.Orbital velocity is the velocity necessary to hold a natural or satellite in its orbit. For a given altitude or wavelength, the more massive the body at the centre of attraction, the greater the orbital velocity.
Complete step by step answer:
Objects that move across the planet in uniform circular motion are considered to be in orbit. This orbit’s velocity depends on the distance between the object and the earth’s centre. This velocity is generally provided to artificial satellites such that every single planet orbits around it.
The formula for orbital velocity is given by-
$v = \sqrt {\dfrac{{GM}}{r}} $, Where $G = $ gravitational constant,$M = $ mass of the body at centre,$r = $ radius of the orbit.
If mass $M$ and radius $R$ are known, the orbital velocity formula is implemented to measure the orbital velocity of any planet. The unit of orbital velocity is meters per second.
From the above expression it is clear that- a satellite’s velocity $v$ varies inversely as the square root of the radius $r$ of the orbit.
So, $v\alpha \dfrac{1}{{\sqrt r }}$ ...... (1)
Given,
Radius of satellite
$A,\,{r_A} = 4R$
Radius of satellite
$B,{r_B} = R$
Speed of satellite
$A,\,{V_A} = 3v$
From equation (1), we can see that-
$
\dfrac{{{V_A}}}{{{V_B}}} = \sqrt {\dfrac{{{r_B}}}{{{r_A}}}} = \sqrt {\dfrac{R}{{4R}}} \\
\Rightarrow \dfrac{{3v}}{{{V_B}}} = \dfrac{1}{2} \\
\therefore{V_B} = 6v \\
$
Hence, the correct option is A.
Additional information:
A satellite’s orbital speed is inversely proportional to the radius of its orbit. If the radius decreases, a satellite’s orbital speed increases. Therefore, the lower altitude satellite has a faster pace.
Note:We may get confused while solving the velocity of satellite B. We should always remember that the velocity is inversely proportional to the radius. So, $\dfrac{{{V_A}}}{{{V_B}}} = \sqrt {\dfrac{{{r_b}}}{{{r_A}}}} $. Hence, ${r_B}$ becomes the numerator, otherwise if we put ${r_A}$ as the numerator we would get a wrong answer. So, we need to be very careful in this step.
Complete step by step answer:
Objects that move across the planet in uniform circular motion are considered to be in orbit. This orbit’s velocity depends on the distance between the object and the earth’s centre. This velocity is generally provided to artificial satellites such that every single planet orbits around it.
The formula for orbital velocity is given by-
$v = \sqrt {\dfrac{{GM}}{r}} $, Where $G = $ gravitational constant,$M = $ mass of the body at centre,$r = $ radius of the orbit.
If mass $M$ and radius $R$ are known, the orbital velocity formula is implemented to measure the orbital velocity of any planet. The unit of orbital velocity is meters per second.
From the above expression it is clear that- a satellite’s velocity $v$ varies inversely as the square root of the radius $r$ of the orbit.
So, $v\alpha \dfrac{1}{{\sqrt r }}$ ...... (1)
Given,
Radius of satellite
$A,\,{r_A} = 4R$
Radius of satellite
$B,{r_B} = R$
Speed of satellite
$A,\,{V_A} = 3v$
From equation (1), we can see that-
$
\dfrac{{{V_A}}}{{{V_B}}} = \sqrt {\dfrac{{{r_B}}}{{{r_A}}}} = \sqrt {\dfrac{R}{{4R}}} \\
\Rightarrow \dfrac{{3v}}{{{V_B}}} = \dfrac{1}{2} \\
\therefore{V_B} = 6v \\
$
Hence, the correct option is A.
Additional information:
A satellite’s orbital speed is inversely proportional to the radius of its orbit. If the radius decreases, a satellite’s orbital speed increases. Therefore, the lower altitude satellite has a faster pace.
Note:We may get confused while solving the velocity of satellite B. We should always remember that the velocity is inversely proportional to the radius. So, $\dfrac{{{V_A}}}{{{V_B}}} = \sqrt {\dfrac{{{r_b}}}{{{r_A}}}} $. Hence, ${r_B}$ becomes the numerator, otherwise if we put ${r_A}$ as the numerator we would get a wrong answer. So, we need to be very careful in this step.
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