The radius of the moon’s orbit about the earth is about $3.6 \times {10^8}m$ The moon’s period is $2.3 \times {10^6}$ seconds. What is the centripetal acceleration of the moon?
Answer
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Hint: In order to solve this question, we will first calculate the velocity of the moon in terms of distance from the earth and its time period and then using values of parameter radius of the orbit and time period we will use general formula of centripetal acceleration of circular motion to calculate the acceleration of the moon.
Formula used:
If v is the velocity of a body following circular motion and r is the radius of circular path then, acceleration a is given by $a = \dfrac{{{v^2}}}{r}$ .
Complete step by step solution:
According to the question, we have given that
$r = 3.6 \times {10^8}m$ radius of circular orbit of the moon around earth.
$T = 2.3 \times {10^6}m$ time period of the moon.
Let v be the velocity of the moon then, total circumference C covered by the moon will be
$C = 2\pi r$ so, velocity can be written as ratio of circumference and time period as
$v = \dfrac{C}{T}$ on putting the value of parameters we get,
$v = \dfrac{{2\pi r}}{T} \to (i)$
now, centripetal acceleration of the moon can be calculated using formula,
$a = \dfrac{{{v^2}}}{r}$ on using the value of v from equation (i) and putting values of each parameters we get,
$a = \dfrac{{{{(\dfrac{{2\pi r}}{T})}^2}}}{r}$
$a = \dfrac{{4{\pi ^2}{r^2}}}{{{T^2}r}}$
$a = \dfrac{{4{{(3.14)}^2} \times (3.6 \times {{10}^8})}}{{{{(2.3 \times {{10}^6})}^2}}}$
on solving, we get
$a = \dfrac{{141.98 \times {{10}^8}}}{{5.29 \times {{10}^{12}}}}$
$a = 26.839 \times {10^{ - 4}}$
or we can write,
$a \approx 0.003m{s^{ - 2}}$
Hence, the centripetal acceleration of the moon is $0.003m{s^{ - 2}}$.
Note:
It should be remembered that the orbit of the moon is assumed to be perfectly circular while in reality the orbit is found to be slightly elliptical and the direction of centripetal acceleration of the moon is always directed towards the centre of the circular path followed by the moon.
Formula used:
If v is the velocity of a body following circular motion and r is the radius of circular path then, acceleration a is given by $a = \dfrac{{{v^2}}}{r}$ .
Complete step by step solution:
According to the question, we have given that
$r = 3.6 \times {10^8}m$ radius of circular orbit of the moon around earth.
$T = 2.3 \times {10^6}m$ time period of the moon.
Let v be the velocity of the moon then, total circumference C covered by the moon will be
$C = 2\pi r$ so, velocity can be written as ratio of circumference and time period as
$v = \dfrac{C}{T}$ on putting the value of parameters we get,
$v = \dfrac{{2\pi r}}{T} \to (i)$
now, centripetal acceleration of the moon can be calculated using formula,
$a = \dfrac{{{v^2}}}{r}$ on using the value of v from equation (i) and putting values of each parameters we get,
$a = \dfrac{{{{(\dfrac{{2\pi r}}{T})}^2}}}{r}$
$a = \dfrac{{4{\pi ^2}{r^2}}}{{{T^2}r}}$
$a = \dfrac{{4{{(3.14)}^2} \times (3.6 \times {{10}^8})}}{{{{(2.3 \times {{10}^6})}^2}}}$
on solving, we get
$a = \dfrac{{141.98 \times {{10}^8}}}{{5.29 \times {{10}^{12}}}}$
$a = 26.839 \times {10^{ - 4}}$
or we can write,
$a \approx 0.003m{s^{ - 2}}$
Hence, the centripetal acceleration of the moon is $0.003m{s^{ - 2}}$.
Note:
It should be remembered that the orbit of the moon is assumed to be perfectly circular while in reality the orbit is found to be slightly elliptical and the direction of centripetal acceleration of the moon is always directed towards the centre of the circular path followed by the moon.
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