The figure shows the motion of a planet around the sun in an elliptical orbit with the sun at the focus. The shaded areas $A$ and $B$ are also shown in the figure,which can be assumed to be equal. If ${t_1}$ and ${t_2}$ represent the time taken for the planet to move from $a$ to $b$ and $d$ to $c$ respectively,then:

  • A
    ${t_1} < {t_2}$
  • B
    ${t_1} > {t_2}$
  • C
    ${t_1} = {t_2}$
  • D
    ${t_1} \le {t_2}$

Explore More

Similar Questions

The time period of a satellite of Earth is $24 \text{ hours}$. If the separation between the Earth and the satellite is decreased to one-fourth of the previous value, then its new time period will become: (in $\text{ hours}$)

The period of revolution increases in the order of:

$A$ planet revolves around the sun in an elliptical orbit. The linear speed of the planet will be maximum at

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 $6 \times 10^{10} \ m$. Mercury will revolve around the sun in:

Difficult
View Solution

$A$ satellite $A$ of mass $m$ is at a distance of $r$ from the centre of the earth. Another satellite $B$ of mass $2m$ is at a distance of $2r$ from the earth's centre. Their time periods are in the ratio of:

Vedclass Products

For Students

Vedclass Test Series

Mock tests in real JEE/NEET style with performance analysis. 5-day free trial.

Start Free Trial
For Teachers

Exam Paper Generator

Generate Set A/B/C/D exam papers from 7.5L+ questions in 2 minutes. 3 chapters free.

Try Free
For Institutes

Online Exam Module

Live online exams with unlimited students, 360° analytics & white-label branding.

See Demo