The figure shows the elliptical path abcd of | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(C) According to Kepler's Second Law,the areal velocity of a planet is constant,meaning the time taken to sweep an area is directly proportional to th

Vedclass · Accepted Answer

$t_1 = 3t_2$

Vedclass · Answer

(C) According to Kepler's Second Law,the areal velocity of a planet is constant,meaning the time taken to sweep an area is directly proportional to the area swept.Let the total area of the ellipse be $A$.The area swept by the planet in path $abc$ is the sum of the area of the semi-ellipse (on the side of $b$) and the area of the triangle $Sca$.Area of semi-ellipse $= \frac{A}{2}$.Area of triangle $Sca = \frac{1}{4}A$ (given).Therefore,the area swept in path $abc$ is $A_1 = \frac{A}{2} + \frac{A}{4} = \frac{3A}{4}$.The area swept in path $cda$ is $A_2 = A - A_1 = A - \frac{3A}{4} = \frac{A}{4}$.Since $\frac{t_1}{t_2} = \frac{A_1}{A_2}$,we have $\frac{t_1}{t_2} = \frac{3A/4}{A/4} = 3$.Thus,$t_1 = 3t_2$.

Similar Questions

The mean radius of the earth's orbit around the sun is $1.5 \times 10^{11} \ m$. The mean radius of the orbit of mercury around the sun is $6 \times 10^{10} \ m$. The mercury will rotate around the sun in

The figure shows an elliptical orbit of a planet of mass $m$ about the sun $S.$ The shaded area $SCD$ is twice the shaded area $SAB.$ If $t_1$ is the time taken for the planet to move from $C$ to $D$ and $t_2$ is the time taken to move from $A$ to $B,$ then:

Which quantity remains constant in the orbital motion of the Earth around the Sun?

The motion of planets in the solar system is an example of the conservation of:

The Earth is revolving around the Sun. If the distance of the Earth from the Sun is reduced to $\frac{1}{4}$ of the present distance,then the length of the present day will be reduced by a factor of:

Vedclass Test Series

Exam Paper Generator

Online Exam Module