VIT - VITEEE 2025
National level exam conducted by VIT University, Vellore | Ranked #11 by NIRF for Engg. | NAAC A++ Accredited | Last Date to Apply: 31st March | NO Further Extensions!
Kepler’s Laws of Planetary Motion is considered one of the most asked concept.
47 Questions around this concept.
Figure shows elliptical path abcd of a planet around the sun $S$ such that the area of triangle csa is $\frac{1}{4}$ the area of the ellipse. (See figure) With db as the semimajor axis, and ca as the semiminor axis. If $t_1$ is the time taken for planet to go over path abc and $t_2$ for path taken over cda then :
The time period of a satellite of earth is 5 hour. If the separation between the earth and the satellite is increased to 4 times the previous value, the new time period will become ( In hours)
"Each planet revolves around the sun in an elliptical orbit with the sun situated at one the foci" is the statement of
JEE Main 2025: City Slip Link | Study Plan | Official Question Paper (Session 1)
JEE Main 2025: Sample Papers | Mock Tests | PYQs | High Scoring Topics | College Predictor
New: Meet Careers360 experts in your city and get guidance on shortlisting colleges
Apply to TOP B.Tech /BE Entrance exams: VITEEE | MET | AEEE | BITSAT
If the radius of earth's orbit is made 1/16, the duration of an year will become
The distance between two planets from the sun is nearly 1014 and 1012 meters. Assuming that they move in circular orbits, their periodic times will be in the ratio.
The figure shows the motion of a planet around the sun in an elliptical orbit with the sun as the focus. The shaded areas A and B are also shown in the figure which can be assumed to be equal. If t1 and t2 represent the time for the planet to move from a to b and d to c respectively, then
If the Earth is one-fourth of its current distance from the Sun, the duration of the year will be
National level exam conducted by VIT University, Vellore | Ranked #11 by NIRF for Engg. | NAAC A++ Accredited | Last Date to Apply: 31st March | NO Further Extensions!
Ranked #42 among Engineering colleges in India by NIRF | Highest CTC 50 LPA , 100% Placements | Last Date to Apply: 28th March
The Earth revolves around the sun in an elliptical orbit with a mean radius of 9.3×107m in 1 year. Assuming that there are no outside influences
If a satellite orbiting the Earth is 9 times closer to the Earth than the Moon, what is the time period of rotation of the satellite? Given rotational time period of Moon $=27$ days and gravitational attraction between the satellite and the moon is neglected.
Kepler gives three empirical laws which govern the motion of the planets which are known as Kepler’s laws of planetary motion.
As we know that planets are large natural bodies rotating around a star in definite orbits.
So, Kepler laws are-
(a) The law of Orbits:
It is Kepler's First Law.
Every planet moves around the sun in an elliptical orbit. And the sun will be at one of the foci of the ellipse.
(b) The law of Area:
It is Kepler's 2nd law.
According to this, the line joining the sun to the planet sweeps out equal areas in equal intervals of time which clearly means that areal velocity is constant. So according to this law, a planet will move slowly when it is farthest from the sun and more rapidly when it is nearest to the sun. You can find it similar to the law of conservation of angular momentum.
For the below figure
Area of velocity $=\frac{d A}{d t}$
$
\frac{d A}{d t}=\frac{1}{2} \frac{(r)(V d t)}{d t}=\frac{1}{2} r V
$
Where
$
\frac{d A}{d t} \rightarrow {\text {Areal velocity }}
$
$d A \rightarrow$ small area traced
Kepler's 2nd law is Similar to the Law of conservation of momentum
As $\frac{d A}{d t}=\frac{L}{2 m}$
where
$
L=m v r \rightarrow \text { Angular momentum }
$
(c) The law of periods:
It is Kepler's 3rd law.
According to this, the square of the Time period of revolutions of any planet around the sun is directly proportional to
the cube of the semi-major axis of that particular orbit.
For the below figure
$
\begin{aligned}
& A B=A F+F B \\
& 2 a=r_1+r_2 \\
& \therefore a=\frac{r_1+r_1}{2}
\end{aligned}
$
Where
$a=$ semi major Axis
$r_1=$ The shortest distance of the planet from the sun (perigee)
$r_2=$ Largest distance of the planet from the sun (apogee)
So if T=Time period of revolution
Then according to Kepler's 3rd law.
$
\begin{aligned}
& T^2 \propto a^3 \\
& \text { or } T^2 \propto\left(\frac{r_1+r_2}{2}\right)^3
\end{aligned}
$
"Stay in the loop. Receive exam news, study resources, and expert advice!"