Gravitation | JEE Main previous year questions with Solutions

STUDY MATERIALS
COURSES
STORE
MORE
SUCCESS STORY
ADMISSION

Back

JEE ADVANCE

CLASS XI - XII

CLASS XI - XII

JEE MAIN

CLASS XI - XII

CLASS XI - XII

NEET

CLASS XI - XII

CLASS XI - XII

BITSAT

CLASS XI - XII

CLASS XI - XII

CBSE BOARD

CLASS IX

CLASS X

CLASS XI

CLASS XII

CLASS VI

CLASS VII

CLASS VIII

UP BOARD

CLASS IX
CLASS X
CLASS XI
CLASS XII

CLASS IX

CLASS X

CLASS XI

CLASS XII

Courses

CLASSROOM COURSES
ONLINE COURSES

ONLINE COURSES

Q 11 :

Two satellites A and B move round the earth in the same orbit. The mass of A is twice the mass of B. The quantity which is same for the two satellites will be [2023]

Potential energy
Total energy
Kinetic energy
Speed

(4)

$P.E. = - \frac{G M_{p} M_{A}}{R}$

$K.E. = + \frac{G M_{p} M_{A}}{2 R}$

$T.E. = - \frac{G M_{p} M_{A}}{2 R}$

$Speed = v = \sqrt{\frac{G M_{p}}{R}}$

$Speed of satellite is independent of mass of satellite.$

Q 12 :

Initially a satellite of 100 kg is in a circular orbit of radius $1.5 R_{E}$ . This satellite can be moved to a circular orbit of radius $3 R_{E}$ by supplying $α \times 10^{6} J$ of energy. The value of $α$ is______.

$($ Take Radius of Earth $R_{E} = 6 \times 10^{6} m and 𝑔 = 10 m / s^{2})$ [2026]

100
500
1000
150

(3)

Energy of a satellite in a circular orbit is given as

$E = \frac{- G M_{E} m}{2 r}; r = radius of circular orbit$

$Required energy to be supplied = Δ E = E_{f} - E_{i}$

$Δ E = (\frac{- G M_{E} m}{2 (3 R_{E})}) - (\frac{- G M_{E} m}{2 (1.5 R_{E})})$

$= \frac{G M_{E} m}{6 R_{E}}$

$Now, g = \frac{G M_{E}}{R_{E}^{2}} \Rightarrow \frac{G M_{E}}{R_{E}} = g R_{E}$

$∴$ $Δ E = \frac{1}{6} g m R_{E}$

$= \frac{1}{6} \times 10 \times 100 \times 6 \times 10^{6}$

$= 1000 \times 10^{6}$

$α = 1000$

Topic Question Set

Q 11 :

Two satellites A and B move round the earth in the same orbit. The mass of A is twice the mass of B. The quantity which is same for the two satellites will be [2023]

Want Us to Email you About Special Offers & Updates?