(D) The angular momentum $L$ of a particle in circular motion is given by $L = mvr$.Since $v = r\omega$ and $\omega = \frac{2\pi}{T}$,we have $L = m(r

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$2.7 \times 10^{40} \ kg \ m^2/s$

(D) The angular momentum $L$ of a particle in circular motion is given by $L = mvr$.Since $v = r\omega$ and $\omega = \frac{2\pi}{T}$,we have $L = m(r\omega)r = m\omega r^2 = m \left( \frac{2\pi}{T} \right) r^2$.Given: $m = 6 \times 10^{24} \ kg$,$r = 1.5 \times 10^8 \ km = 1.5 \times 10^{11} \ m$,$T = 3.14 \times 10^7 \ s$.Substituting the values:$L = \frac{6 \times 10^{24} \times 2 \times 3.14 \times (1.5 \times 10^{11})^2}{3.14 \times 10^7}$$L = \frac{6 \times 10^{24} \times 2 \times 2.25 \times 10^{22}}{10^7}$$L = 27 \times 10^{39} = 2.7 \times 10^{40} \ kg \ m^2/s$.

Class 11 Physics Gravitation Motion of Satellites in Circular Orbits and Planets in Elliptical Orbits

Medium

If the Earth is considered a point mass of $6 \times 10^{24} \ kg$ revolving around the Sun at a distance of $1.5 \times 10^8 \ km$ in a period of $T = 3.14 \times 10^7 \ s$,then the angular momentum of the Earth around the Sun will be:

A
$1.2 \times 10^{18} \ kg \ m^2/s$
B
$1.8 \times 10^{29} \ kg \ m^2/s$
C
$1.5 \times 10^{37} \ kg \ m^2/s$
D
$2.7 \times 10^{40} \ kg \ m^2/s$

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Gravitation

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Motion of Satellites in Circular Orbits and Planets in Elliptical Orbits

Which force is responsible for the necessary centripetal force of a satellite revolving around the Earth?

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The mean radius of the earth is $R$,its angular speed on its own axis is $\omega$,and the acceleration due to gravity at the earth's surface is $g$. The cube of the radius of the orbit of a geostationary satellite will be

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$A$ planet is revolving around the sun in a circular orbit with a radius $r$. The time period is $T$. If the force between the planet and the star is proportional to $r^{-3/2}$,then the square of the time period is proportional to

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$A$ planet moving along an elliptical orbit is closest to the sun at a distance $r_1$ and farthest away at a distance of $r_2$. If $v_1$ and $v_2$ are the linear velocities at these points respectively,then the ratio $\frac{v_1}{v_2}$ is

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$A$ planet revolves around the sun in an elliptical orbit. When it is closest to the sun,its distance from the sun is $d_1$ and its velocity is $v_1$. When it is farthest from the sun,its distance from the sun is $d_2$ and its velocity is $v_2$. Find the value of $v_2$.

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If the Earth is considered a point mass of $6 \times 10^{24} \ kg$ revolving around the Sun at a distance of $1.5 \times 10^8 \ km$ in a period of $T = 3.14 \times 10^7 \ s$,then the angular momentum of the Earth around the Sun will be:

Similar Questions

Which force is responsible for the necessary centripetal force of a satellite revolving around the Earth?

The mean radius of the earth is $R$,its angular speed on its own axis is $\omega$,and the acceleration due to gravity at the earth's surface is $g$. The cube of the radius of the orbit of a geostationary satellite will be

$A$ planet is revolving around the sun in a circular orbit with a radius $r$. The time period is $T$. If the force between the planet and the star is proportional to $r^{-3/2}$,then the square of the time period is proportional to

$A$ planet moving along an elliptical orbit is closest to the sun at a distance $r_1$ and farthest away at a distance of $r_2$. If $v_1$ and $v_2$ are the linear velocities at these points respectively,then the ratio $\frac{v_1}{v_2}$ is

$A$ planet revolves around the sun in an elliptical orbit. When it is closest to the sun,its distance from the sun is $d_1$ and its velocity is $v_1$. When it is farthest from the sun,its distance from the sun is $d_2$ and its velocity is $v_2$. Find the value of $v_2$.

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