Question

The ratio of Earth’s orbital momentum (about the Sun) to its mass is $4.4 \times {10^{15}}{m^2}/s$. Find the area enclosed by Earth’s circular orbit. (Given 1 year= $3.15 \times {10^7}$ second)
A. ${\text{6}}{{.93 \times 1}}{{\text{0}}^{{\text{22}}}}{\text{c}}{{\text{m}}^{\text{2}}}$
B. ${\text{6}}{{.93 \times 1}}{{\text{0}}^{11}}{{\text{m}}^{\text{2}}}$
C. ${\text{6}}{{.93 \times 1}}{{\text{0}}^{22}}{{\text{m}}^{\text{2}}}$
D. ${\text{6}}{{.93 \times 1}}{{\text{0}}^{ - 22}}{{\text{m}}^{\text{2}}}$

Answer 1

Hint: We are looking for the areal velocity of a planet around the Sun. Also, we need to do some integration after considering a small part.

Formula used: To calculate the areal velocity of a planet around the Sun:
\[\dfrac{{{\text{dA}}}}{{{\text{dT}}}}{\text{ = }}\dfrac{{\text{L}}}{{{\text{2m}}}}\]
Here, \[{\text{L}}\] is the angular momentum of the planet about the Sun.
\[{\text{m}}\] is the mass of the planet

Complete step by step answer:
It is already known that the areal velocity of a planet around the Sun= \[\dfrac{{{\text{dA}}}}{{{\text{dT}}}}{\text{ = }}\dfrac{{\text{L}}}{{{\text{2m}}}}\]
On integrating on both sides,
$\int {{\text{dA = }}\dfrac{{\text{L}}}{{{\text{2m}}}}\int {{\text{dt}}} } $
$ \Rightarrow {\text{A = }}\dfrac{{\text{L}}}{{{\text{2m}}}}{\text{T}}$
Putting all the given values,
${\text{A = }}\dfrac{{\text{1}}}{{\text{2}}}{{ \times 4}}{{.4 \times 1}}{{\text{0}}^{{\text{15}}}}{{ \times 365 \times 24 \times 3600}}{{\text{m}}^{\text{2}}}$
$ \Rightarrow {\text{A = 6}}{{.93 \times 1}}{{\text{0}}^{{\text{22}}}}{{\text{m}}^{\text{2}}}$
So, we need to see from the above options, and select the correct value.
Thus, the correct answer is option A.
Additional Note:
When a particle moves along a curve, the rate of change of area swept by the particle is known as areal velocity. It is defined in respect to planetary motion. Every particle in the solar system has a real velocity. The Sun, The Earth and all the Planets. The areal velocity of earth is perpendicular to the area swept by the earth in the given time interval around the sun in her orbital path. It is normal to the plane containing the earth and the sun as areal velocity.

Note: Areal velocity means the rate at which area is swept by a particle as it moves around a curve. Also, the orbital angular momentum of an object about a chosen origin is defined as the angular momentum of the centre of mass about the origin. The total angular momentum of an object is the sum of the spin and orbital angular momentum.