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Question: Assuming that the earth's orbit around the sun is circular, find the linear velocity of its motion a...

Assuming that the earth's orbit around the sun is circular, find the linear velocity of its motion and the period of revolution about the sun. Given G=6.67×10−11G = 6.67 \times {10^{ - 11}} S.I. units, mass of the sun=1.99×10−30kgsun = 1.99 \times {10^{ - 30}}kg, mean distance between the sun and the earth=1.497×10−11mearth = 1.497 \times {10^{ - 11}}m

Explanation

Solution

Hint:- Let the radius of the orbit be a and the speed of the planet in the orbit is v. by Newton's law , the force on the planet equals its mass times the acceleration .thus,
Mathematically GMma2=m(v2)a or v = GMa  \dfrac{{GMm}}{{{a^2}}} = \dfrac{{m({v^2})}}{a} \\\ or{\text{ v = }}\sqrt {\dfrac{{GM}}{a}} \\\
This is the linear speed of the planet to revolve in the orbit .

Complete step-by-step solution :As we all know that linear velocity is the velocity of the planet to rotate around the sun.
Mathematically v=GMrv = \sqrt {\dfrac{{GM}}{r}} where v=linear velocity
G is gravitational constant
M is mass of sun
r is the mean distance of the planet from earth.
Now we have :-
G=6.67×10−11G = 6.67 \times {10^{ - 11}} S.I. unit
r=1.497×10−11mr = 1.497 \times {10^{ - 11}}m
M=1.99×10−30kgM = 1.99 \times {10^{ - 30}}kg
Putting all values in the linear velocity equation.
v=GMr v=6.67×10−11×1.99×10301.497×10−11  v = \sqrt {\dfrac{{GM}}{r}} \\\ v = \sqrt {\dfrac{{6.67 \times {{10}^{ - 11}} \times 1.99 \times {{10}^{30}}}}{{1.497 \times {{10}^{ - 11}}}}} \\\
On solving above equation:
v=8.86×1030 v=2.97×1015 ms−1  v = \sqrt {8.86 \times {{10}^{30}}} \\\ v = 2.97 \times {10^{15}}{\text{ m}}{{\text{s}}^{ - 1}} \\\
∴\therefore this velocity is the linear velocity
So we can write :
v=ωrv = \omega rwhere ω\omega is angular velocity
And r is radius of circle
∴ω=2πT\therefore \omega = \dfrac{{2\pi }}{T}here T is time of revolution
T=2πω T=2πrv  T = \dfrac{{2\pi }}{\omega } \\\ T = \dfrac{{2\pi r}}{v} \\\
On putting the all variables values
T=2×3.14×1.497×10−112.97×1015 T=3.165×10−20 s  T = \dfrac{{2 \times 3.14 \times 1.497 \times {{10}^{ - 11}}}}{{2.97 \times {{10}^{15}}}} \\\ T = 3.165 \times {10^{ - 20{\text{ }}}}s \\\
Hence value of linear velocity is v=2.97×1015 ms−1v = 2.97 \times {10^{15}}{\text{ m}}{{\text{s}}^{ - 1}}
And value of time of revolution is T=3.165×10−20 sec⁡ondT = 3.165 \times {10^{ - 20{\text{ }}}}\sec ond

Note:- Planets move around the sun due to gravitational attraction of the sun. The path of these planets are elliptical with the sun at focus. However the difference in the major and minor axes is not large. The orbits can be treated as nearly circular for not too sophisticated calculations.