
An artificial satellite is moving in a circular orbit of radius 42250km. Calculate the speed it takes 24 hours to revolve around the earth.
$\text{A}\text{. }3.07km{{s}^{-1}}$
$\text{B}\text{. }5km{{s}^{-1}}$
$\text{C}\text{. 4}.5km{{s}^{-1}}$
$\text{D}\text{. }7km{{s}^{-1}}$
Answer
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Hint: The relation between the time period of the satellite and its angular velocity (also called angular frequency) is given as $T=\dfrac{2\pi }{\omega }$. The velocity of the satellite is given as $v=r\omega $, where r is the radius of the circular path of the satellite.
Complete step-by-step answer:
Objects revolving in circular orbits under the influence of a larger planet or star, have a constant speed. You may know that when a body is in a circular motion with a constant speed, then its motion is called a uniform circular motion.
A uniform circular motion is a periodic motion. Therefore, it has a time period of its motion. Here, the satellite has a time period of revolution.
The relation between the time period of the satellite and its angular velocity (also called angular frequency) is given as $T=\dfrac{2\pi }{\omega }\Rightarrow \omega =\dfrac{2\pi }{T}$ …….. (i).
The velocity of the satellite is given as $v=r\omega $ …….. (ii).
r in equation (ii) is the radius of the circular path of the satellite.
Substitute the value of angular velocity $\omega $ from equation (i) in equation (ii).
Therefore, we get that
$v=r.\dfrac{2\pi }{T}$ ….. (iii).
In this case, r = 42250 km, T = 24 hours.
Before that, we have to calculate the time period in the units of seconds.
We know that one hour is equal to 60 minutes and one minute is equal to 60 seconds.
Therefore, $1hr=60\times 60s=3600s$.
This implies that T = $24\times 3600s$.
Substitute the values of r and T in equation (iii).
Therefore, $v=(42250).\dfrac{2\pi }{(24\times 3600)}=3.07m{{s}^{-1}}$.
The speed of the satellite is 3.07$m{{s}^{-1}}$.
Hence, the correct option is A.
Note: Angular velocity and the angular frequency of same quantities. When we deal with waves, we call it angular frequency and when we deal with rotational mechanics, we call it angular velocity.
Complete step-by-step answer:
Objects revolving in circular orbits under the influence of a larger planet or star, have a constant speed. You may know that when a body is in a circular motion with a constant speed, then its motion is called a uniform circular motion.
A uniform circular motion is a periodic motion. Therefore, it has a time period of its motion. Here, the satellite has a time period of revolution.
The relation between the time period of the satellite and its angular velocity (also called angular frequency) is given as $T=\dfrac{2\pi }{\omega }\Rightarrow \omega =\dfrac{2\pi }{T}$ …….. (i).
The velocity of the satellite is given as $v=r\omega $ …….. (ii).
r in equation (ii) is the radius of the circular path of the satellite.
Substitute the value of angular velocity $\omega $ from equation (i) in equation (ii).
Therefore, we get that
$v=r.\dfrac{2\pi }{T}$ ….. (iii).
In this case, r = 42250 km, T = 24 hours.
Before that, we have to calculate the time period in the units of seconds.
We know that one hour is equal to 60 minutes and one minute is equal to 60 seconds.
Therefore, $1hr=60\times 60s=3600s$.
This implies that T = $24\times 3600s$.
Substitute the values of r and T in equation (iii).
Therefore, $v=(42250).\dfrac{2\pi }{(24\times 3600)}=3.07m{{s}^{-1}}$.
The speed of the satellite is 3.07$m{{s}^{-1}}$.
Hence, the correct option is A.
Note: Angular velocity and the angular frequency of same quantities. When we deal with waves, we call it angular frequency and when we deal with rotational mechanics, we call it angular velocity.
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