Question:
India’s Mangalyan was sent to the Mars by launching it into a transfer orbit EOM around the sun. It leaves the earth at $\mathrm{E}$ and meets Mars at $\mathrm{M}$. If the semi-major axis of Earth’s orbit is $\mathrm{a}_{\mathrm{e}}=1.5 \times 10^{11} \mathrm{~m}$, that of Mars orbit $\mathrm{a}_{\mathrm{m}}=$ $2.28 \times 10^{11} \mathrm{~m}$, taken Kepler’s laws give the estimate of time for Mangalyan to reach Mars from Earth to be close to:
500 days
320 days
260 days
220 days
Question of from chapter.
JEE Main Previous Year April 9, 2014
Correct Option: 2
Solution:
(2)
Download now India’s Best Exam Preparation App
Class 9-10, JEE & NEET
Related Questions
If the angular momentum of a planet of mass $\mathrm{m}$, moving around the Sun in a circular orbit is $\mathrm{L}$, about the center of the Sun, its areal velocity is:
Figure shows elliptical path abed 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 $\mathrm{t}_{1}$ is the time taken for planet to go over path $a b c$ and $t_{2}$ for path taken over cda then:
The time period of a satellite of earth is 5 hours. If the separation between the earth and the satellite is increased to 4 times the previous value, the new time period will become
A straight rod of length $\mathrm{L}$ extends from $x=a$ to $x=\mathrm{L}+\mathrm{a}$.
The gravitational force it exerts on point mass ‘ $\mathrm{m}$ ‘ at $x=0$,
if the mass per unit length of the rod is $\mathrm{A}+\mathrm{B} x^{2}$, is given
by:
Take the mean distance of the moon and the sun from the earth to be $0.4 \times 10^{6} \mathrm{~km}$ and $150 \times 10^{6} \mathrm{~km}$ respectively. Their masses are $8 \times 10^{22} \mathrm{~kg}$ and $2 \times 10^{30} \mathrm{~kg}$ respectively. The radius of the earth is $6400 \mathrm{~km}$. Let $\Delta \mathrm{F}_{1}$ be the difference in the forces exerted by the moon at the nearest and farthest points on the earth and $\Delta \mathrm{F}_{2}$ be the difference in the force exerted by the sun at the nearest and farthest points on the earth. Then, the number closest to $\frac{\Delta F_{1}}{\Delta F_{2}}$ is:
Four particles, each of mass $M$ and equidistant from each other, move along a circle of radius $R$ under the action of their mutual gravitational attraction. The speed of each particle is:
From a sphere of mass $M$ and radius $R$, a smaller sphere of radius $\frac{R}{2}$ is carved out such that the cavity made in the original sphere is between its centre and the periphery (See figure). For the configuration in the figure where the distance between the centre of the original sphere and the removed sphere is $3 R$, the gravitational force between the two sphere is:
Two particles of equal mass ‘ $m$ ‘ go around a circle of radius $R$ under the action of their mutual gravitational attraction. The speed of each particle with respect to their centre of mass is
Two spherical bodies of mass $M$ and $5 M \&$ radii $R \& 2 R$ respectively are released in free space with initial separation between their centres equal to $12 R$. If they attract each other due to gravitational force only, then the distance covered by the smaller body just before collision is
The value of acceleration due to gravity is $g_{1}$ at a height $h=\frac{R}{2}(R=$ radius of the earth $)$ from the surface of the earth. It is again equal to $g_{1}$ and a depth $d$ below the surface of the earth. The ratio $\left(\frac{d}{R}\right)$ equals :
