JEE Main Previous Year Questions (2025): Gravitation

 Page 1


JEE Main Previous Year Qs (2025): 
Gravitation 
Q1: A satellite of mass 
?? ?? is revolving around earth in a circular orbit at a height of 
?? ?? from 
earth surface. The angular momentum of the satellite is ?? v
?????? ?? . The value of ?? is ____ , where 
?? and ?? are the mass and radius of earth, respectively. ( ?? is the gravitational constant) 
JEE Main 2025 (Online) 23rd January Evening Shift 
Ans: 3 
Solution: 
?? = ?? +
?? 3
=
4?? 3
 
For a circular orbit, the gravitational force provides the required centripetal force: 
?????? ?? 2
=
?? ?? 2
?? 
which simplifies to 
?? = v
????
?? . 
The angular momentum ?? of the satellite is given by 
?? = ?????? . 
Substitute the values: 
Satellite mass: ?? =
?? 2
 
Orbital radius: ?? =
4?? 3
 
Speed: ?? =
v
????
4?? 3
= v
3????
4?? 
Thus, 
?? =
?? 2
·
4?? 3
· v
3????
4?? . 
Simplify the expression: 
?? =
?? ·4?? 6
v
3????
4?? =
2????
3
v
3????
4?? . 
Notice that the square root can be combined as: 
v
3????
4?? = v
3
4
v
????
?? . 
Therefore, 
?? =
2????
3
· v
3
4
v
????
?? =
2???? v 3
3·2
v
????
?? =
?? v 3
3
v ?????? , 
where we used the fact that ?? v
1
?? = v ?? to form v ?????? . 
Page 2


JEE Main Previous Year Qs (2025): 
Gravitation 
Q1: A satellite of mass 
?? ?? is revolving around earth in a circular orbit at a height of 
?? ?? from 
earth surface. The angular momentum of the satellite is ?? v
?????? ?? . The value of ?? is ____ , where 
?? and ?? are the mass and radius of earth, respectively. ( ?? is the gravitational constant) 
JEE Main 2025 (Online) 23rd January Evening Shift 
Ans: 3 
Solution: 
?? = ?? +
?? 3
=
4?? 3
 
For a circular orbit, the gravitational force provides the required centripetal force: 
?????? ?? 2
=
?? ?? 2
?? 
which simplifies to 
?? = v
????
?? . 
The angular momentum ?? of the satellite is given by 
?? = ?????? . 
Substitute the values: 
Satellite mass: ?? =
?? 2
 
Orbital radius: ?? =
4?? 3
 
Speed: ?? =
v
????
4?? 3
= v
3????
4?? 
Thus, 
?? =
?? 2
·
4?? 3
· v
3????
4?? . 
Simplify the expression: 
?? =
?? ·4?? 6
v
3????
4?? =
2????
3
v
3????
4?? . 
Notice that the square root can be combined as: 
v
3????
4?? = v
3
4
v
????
?? . 
Therefore, 
?? =
2????
3
· v
3
4
v
????
?? =
2???? v 3
3·2
v
????
?? =
?? v 3
3
v ?????? , 
where we used the fact that ?? v
1
?? = v ?? to form v ?????? . 
So the final expression is: 
?? =
?? v 3
v ?????? . 
It is given that the angular momentum can also be expressed as: 
?? = ?? v
?????? ?? . 
Equate the two expressions: 
?? v 3
v ?????? = ?? v
?????? ?? . 
Cancel ?? and v ?????? on both sides (assuming they are nonzero): 
1
v 3
= v
1
?? . 
Taking squares on both sides: 
1
3
=
1
?? . 
Thus, 
?? = 3. 
Q2: Acceleration due to gravity on the surface of earth is ' ?? ' . If the diameter of earth is 
reduced to one third of its original value and mass remains unchanged, then the acceleration 
due to gravity on the surface of the earth is ____ g. 
JEE Main 2025 (Online) 24th January Evening Shift 
Ans: 9 
Solution: 
? acceleration due to gravity on surface is given by 
?? =
????
?? ?? 2
 
Now since diameter is reduced to 1/3
rd 
, radius also reduces to 1/3
rd 
, keeping mass constant 
New value of acceleration due to gravity on Earth's surface is 
?? '
=
GM
(
R
e
3
)
2
= 9
GMe
R
e
2
= 9 g 
Q3: Two planets, ?? and ?? are orbiting a common star in circular orbits of radii ?? ?? and ?? ?? , 
respectively, with ?? ?? = ?? ?? ?? . The planet ?? is ?? v ?? times more massive than planet ?? . The 
ratio (
?? ?? ?? ?? ) of angular momentum ( ?? ?? ) of planet ?? to that of planet ?? ( ?? ?? ) is closest to integer 
____ . 
JEE Main 2025 (Online) 29th January Evening Shift 
Ans: 8 
Solution: 
Page 3


JEE Main Previous Year Qs (2025): 
Gravitation 
Q1: A satellite of mass 
?? ?? is revolving around earth in a circular orbit at a height of 
?? ?? from 
earth surface. The angular momentum of the satellite is ?? v
?????? ?? . The value of ?? is ____ , where 
?? and ?? are the mass and radius of earth, respectively. ( ?? is the gravitational constant) 
JEE Main 2025 (Online) 23rd January Evening Shift 
Ans: 3 
Solution: 
?? = ?? +
?? 3
=
4?? 3
 
For a circular orbit, the gravitational force provides the required centripetal force: 
?????? ?? 2
=
?? ?? 2
?? 
which simplifies to 
?? = v
????
?? . 
The angular momentum ?? of the satellite is given by 
?? = ?????? . 
Substitute the values: 
Satellite mass: ?? =
?? 2
 
Orbital radius: ?? =
4?? 3
 
Speed: ?? =
v
????
4?? 3
= v
3????
4?? 
Thus, 
?? =
?? 2
·
4?? 3
· v
3????
4?? . 
Simplify the expression: 
?? =
?? ·4?? 6
v
3????
4?? =
2????
3
v
3????
4?? . 
Notice that the square root can be combined as: 
v
3????
4?? = v
3
4
v
????
?? . 
Therefore, 
?? =
2????
3
· v
3
4
v
????
?? =
2???? v 3
3·2
v
????
?? =
?? v 3
3
v ?????? , 
where we used the fact that ?? v
1
?? = v ?? to form v ?????? . 
So the final expression is: 
?? =
?? v 3
v ?????? . 
It is given that the angular momentum can also be expressed as: 
?? = ?? v
?????? ?? . 
Equate the two expressions: 
?? v 3
v ?????? = ?? v
?????? ?? . 
Cancel ?? and v ?????? on both sides (assuming they are nonzero): 
1
v 3
= v
1
?? . 
Taking squares on both sides: 
1
3
=
1
?? . 
Thus, 
?? = 3. 
Q2: Acceleration due to gravity on the surface of earth is ' ?? ' . If the diameter of earth is 
reduced to one third of its original value and mass remains unchanged, then the acceleration 
due to gravity on the surface of the earth is ____ g. 
JEE Main 2025 (Online) 24th January Evening Shift 
Ans: 9 
Solution: 
? acceleration due to gravity on surface is given by 
?? =
????
?? ?? 2
 
Now since diameter is reduced to 1/3
rd 
, radius also reduces to 1/3
rd 
, keeping mass constant 
New value of acceleration due to gravity on Earth's surface is 
?? '
=
GM
(
R
e
3
)
2
= 9
GMe
R
e
2
= 9 g 
Q3: Two planets, ?? and ?? are orbiting a common star in circular orbits of radii ?? ?? and ?? ?? , 
respectively, with ?? ?? = ?? ?? ?? . The planet ?? is ?? v ?? times more massive than planet ?? . The 
ratio (
?? ?? ?? ?? ) of angular momentum ( ?? ?? ) of planet ?? to that of planet ?? ( ?? ?? ) is closest to integer 
____ . 
JEE Main 2025 (Online) 29th January Evening Shift 
Ans: 8 
Solution: 
Let a planet of mass ?? orbits a star of mass ?? in circular orbit of radius ?? with speed ?? . 
 
For this circular motion, centripetal force is provided by the gravitational force between both masses. 
Hence, ?? ?? = ?? 4
 
?
?? ?? 2
?? =
?????? ?? 2
 
? ?? =
v
????
?? 
Angular momentum of the planet, ?? = ?????? 
? ?? = ?? v
????
?? ?? 
? ?? = v ???? ?? v ?? 
Given, ?? ?? = 4v 2?? ?? and ?? ?? = 2?? ?? 
M is same for both. 
So, 
?? ?? ?? ?? = (
?? ?? ?? ?? )v
?? ?? ?? ?? 
= ( 4v 2) v 2 = 4 × 2 
?
?? ?? ?? ?? = 8. 
 
Q4: A satellite of mass ???????? ???? is launched to revolve around the earth in an orbit at a 
height of 270 km from the earth's surface. Kinetic energy of the satellite in this orbit is ____ ×
????
????
 ?? . 
(Mass of earth = ?? × ????
????
 ???? , Radius of earth = ?? . ?? × ????
?? ?? , Gravitational constant =
?? . ???? × ?? ?? -????
????
?? ????
-?? ) 
JEE Main 2025 (Online) 2nd April Evening Shift 
Page 4


JEE Main Previous Year Qs (2025): 
Gravitation 
Q1: A satellite of mass 
?? ?? is revolving around earth in a circular orbit at a height of 
?? ?? from 
earth surface. The angular momentum of the satellite is ?? v
?????? ?? . The value of ?? is ____ , where 
?? and ?? are the mass and radius of earth, respectively. ( ?? is the gravitational constant) 
JEE Main 2025 (Online) 23rd January Evening Shift 
Ans: 3 
Solution: 
?? = ?? +
?? 3
=
4?? 3
 
For a circular orbit, the gravitational force provides the required centripetal force: 
?????? ?? 2
=
?? ?? 2
?? 
which simplifies to 
?? = v
????
?? . 
The angular momentum ?? of the satellite is given by 
?? = ?????? . 
Substitute the values: 
Satellite mass: ?? =
?? 2
 
Orbital radius: ?? =
4?? 3
 
Speed: ?? =
v
????
4?? 3
= v
3????
4?? 
Thus, 
?? =
?? 2
·
4?? 3
· v
3????
4?? . 
Simplify the expression: 
?? =
?? ·4?? 6
v
3????
4?? =
2????
3
v
3????
4?? . 
Notice that the square root can be combined as: 
v
3????
4?? = v
3
4
v
????
?? . 
Therefore, 
?? =
2????
3
· v
3
4
v
????
?? =
2???? v 3
3·2
v
????
?? =
?? v 3
3
v ?????? , 
where we used the fact that ?? v
1
?? = v ?? to form v ?????? . 
So the final expression is: 
?? =
?? v 3
v ?????? . 
It is given that the angular momentum can also be expressed as: 
?? = ?? v
?????? ?? . 
Equate the two expressions: 
?? v 3
v ?????? = ?? v
?????? ?? . 
Cancel ?? and v ?????? on both sides (assuming they are nonzero): 
1
v 3
= v
1
?? . 
Taking squares on both sides: 
1
3
=
1
?? . 
Thus, 
?? = 3. 
Q2: Acceleration due to gravity on the surface of earth is ' ?? ' . If the diameter of earth is 
reduced to one third of its original value and mass remains unchanged, then the acceleration 
due to gravity on the surface of the earth is ____ g. 
JEE Main 2025 (Online) 24th January Evening Shift 
Ans: 9 
Solution: 
? acceleration due to gravity on surface is given by 
?? =
????
?? ?? 2
 
Now since diameter is reduced to 1/3
rd 
, radius also reduces to 1/3
rd 
, keeping mass constant 
New value of acceleration due to gravity on Earth's surface is 
?? '
=
GM
(
R
e
3
)
2
= 9
GMe
R
e
2
= 9 g 
Q3: Two planets, ?? and ?? are orbiting a common star in circular orbits of radii ?? ?? and ?? ?? , 
respectively, with ?? ?? = ?? ?? ?? . The planet ?? is ?? v ?? times more massive than planet ?? . The 
ratio (
?? ?? ?? ?? ) of angular momentum ( ?? ?? ) of planet ?? to that of planet ?? ( ?? ?? ) is closest to integer 
____ . 
JEE Main 2025 (Online) 29th January Evening Shift 
Ans: 8 
Solution: 
Let a planet of mass ?? orbits a star of mass ?? in circular orbit of radius ?? with speed ?? . 
 
For this circular motion, centripetal force is provided by the gravitational force between both masses. 
Hence, ?? ?? = ?? 4
 
?
?? ?? 2
?? =
?????? ?? 2
 
? ?? =
v
????
?? 
Angular momentum of the planet, ?? = ?????? 
? ?? = ?? v
????
?? ?? 
? ?? = v ???? ?? v ?? 
Given, ?? ?? = 4v 2?? ?? and ?? ?? = 2?? ?? 
M is same for both. 
So, 
?? ?? ?? ?? = (
?? ?? ?? ?? )v
?? ?? ?? ?? 
= ( 4v 2) v 2 = 4 × 2 
?
?? ?? ?? ?? = 8. 
 
Q4: A satellite of mass ???????? ???? is launched to revolve around the earth in an orbit at a 
height of 270 km from the earth's surface. Kinetic energy of the satellite in this orbit is ____ ×
????
????
 ?? . 
(Mass of earth = ?? × ????
????
 ???? , Radius of earth = ?? . ?? × ????
?? ?? , Gravitational constant =
?? . ???? × ?? ?? -????
????
?? ????
-?? ) 
JEE Main 2025 (Online) 2nd April Evening Shift 
Ans: 3 
Solution: 
The kinetic energy (KE) of a satellite revolving around the Earth can be expressed with the following 
formula: 
KE =
1
2
?? ?? 2
=
1
2
?? ?? ?? ?? ?? =
?? ?? ?? ?? 2?? 
For this satellite, the radius ?? is the sum of the Earth's radius ?? ?? and the height h above the Earth's 
surface: 
?? = ?? ?? + h 
Given: 
Mass of the satellite, ?? = 1000 kg 
Mass of the Earth, ?? ?? = 6 × 10
24
 kg 
Radius of the Earth, ?? ?? = 6.4 × 10
6
 m 
Height of orbit above the Earth's surface, h = 270 km = 2.7 × 10
5
 m 
Gravitational constant, ?? = 6.67 × 10
-11
Nm
2
 kg
-2
 
Substitute these values into the equation: 
KE =
6.67 × 10
-11
× 6 × 10
24
× 1000
2( 6.4 × 10
6
+ 2.7 × 10
5
)
 
=
6.67 × 10
-11
× 6 × 10
24
× 1000
2 × 6.67 × 10
6
 
= 3 × 10
10
 J 
Hence, the kinetic energy of the satellite in its orbit is 3 × 10
10
 J. 
Q5: Three identical spheres of mass ?? , are placed at the vertices of an equilateral triangle of 
length ?? . When released, they interact only through gravitational force and collide after a 
time ?? = ?? seconds. If the sides of the triangle are increased to length ?? ?? and also the 
masses of the spheres are made 2 m , then they will collide after ____ seconds. 
JEE Main 2025 (Online) 3rd April Morning Shift 
Ans: 8 
Solution: 
Dimensional Analysis: 
?? ? ?? ?? ?? ?? ?? ?? 
Using dimensional analysis for gravitational interactions, we have: 
T ? M
x
?? y
?? z
 
Where ?? has dimensions [M
-1
 L
3
 T
-2
]. 
Solving for Exponents: 
T ? M
x-y
L
3?? +?? T
-2?? 
Equating dimensions, we solve: 
x - y = 0 ? x = y 
-2y = 1 ? y = -
1
2
, x = -
1
2
 
Page 5


JEE Main Previous Year Qs (2025): 
Gravitation 
Q1: A satellite of mass 
?? ?? is revolving around earth in a circular orbit at a height of 
?? ?? from 
earth surface. The angular momentum of the satellite is ?? v
?????? ?? . The value of ?? is ____ , where 
?? and ?? are the mass and radius of earth, respectively. ( ?? is the gravitational constant) 
JEE Main 2025 (Online) 23rd January Evening Shift 
Ans: 3 
Solution: 
?? = ?? +
?? 3
=
4?? 3
 
For a circular orbit, the gravitational force provides the required centripetal force: 
?????? ?? 2
=
?? ?? 2
?? 
which simplifies to 
?? = v
????
?? . 
The angular momentum ?? of the satellite is given by 
?? = ?????? . 
Substitute the values: 
Satellite mass: ?? =
?? 2
 
Orbital radius: ?? =
4?? 3
 
Speed: ?? =
v
????
4?? 3
= v
3????
4?? 
Thus, 
?? =
?? 2
·
4?? 3
· v
3????
4?? . 
Simplify the expression: 
?? =
?? ·4?? 6
v
3????
4?? =
2????
3
v
3????
4?? . 
Notice that the square root can be combined as: 
v
3????
4?? = v
3
4
v
????
?? . 
Therefore, 
?? =
2????
3
· v
3
4
v
????
?? =
2???? v 3
3·2
v
????
?? =
?? v 3
3
v ?????? , 
where we used the fact that ?? v
1
?? = v ?? to form v ?????? . 
So the final expression is: 
?? =
?? v 3
v ?????? . 
It is given that the angular momentum can also be expressed as: 
?? = ?? v
?????? ?? . 
Equate the two expressions: 
?? v 3
v ?????? = ?? v
?????? ?? . 
Cancel ?? and v ?????? on both sides (assuming they are nonzero): 
1
v 3
= v
1
?? . 
Taking squares on both sides: 
1
3
=
1
?? . 
Thus, 
?? = 3. 
Q2: Acceleration due to gravity on the surface of earth is ' ?? ' . If the diameter of earth is 
reduced to one third of its original value and mass remains unchanged, then the acceleration 
due to gravity on the surface of the earth is ____ g. 
JEE Main 2025 (Online) 24th January Evening Shift 
Ans: 9 
Solution: 
? acceleration due to gravity on surface is given by 
?? =
????
?? ?? 2
 
Now since diameter is reduced to 1/3
rd 
, radius also reduces to 1/3
rd 
, keeping mass constant 
New value of acceleration due to gravity on Earth's surface is 
?? '
=
GM
(
R
e
3
)
2
= 9
GMe
R
e
2
= 9 g 
Q3: Two planets, ?? and ?? are orbiting a common star in circular orbits of radii ?? ?? and ?? ?? , 
respectively, with ?? ?? = ?? ?? ?? . The planet ?? is ?? v ?? times more massive than planet ?? . The 
ratio (
?? ?? ?? ?? ) of angular momentum ( ?? ?? ) of planet ?? to that of planet ?? ( ?? ?? ) is closest to integer 
____ . 
JEE Main 2025 (Online) 29th January Evening Shift 
Ans: 8 
Solution: 
Let a planet of mass ?? orbits a star of mass ?? in circular orbit of radius ?? with speed ?? . 
 
For this circular motion, centripetal force is provided by the gravitational force between both masses. 
Hence, ?? ?? = ?? 4
 
?
?? ?? 2
?? =
?????? ?? 2
 
? ?? =
v
????
?? 
Angular momentum of the planet, ?? = ?????? 
? ?? = ?? v
????
?? ?? 
? ?? = v ???? ?? v ?? 
Given, ?? ?? = 4v 2?? ?? and ?? ?? = 2?? ?? 
M is same for both. 
So, 
?? ?? ?? ?? = (
?? ?? ?? ?? )v
?? ?? ?? ?? 
= ( 4v 2) v 2 = 4 × 2 
?
?? ?? ?? ?? = 8. 
 
Q4: A satellite of mass ???????? ???? is launched to revolve around the earth in an orbit at a 
height of 270 km from the earth's surface. Kinetic energy of the satellite in this orbit is ____ ×
????
????
 ?? . 
(Mass of earth = ?? × ????
????
 ???? , Radius of earth = ?? . ?? × ????
?? ?? , Gravitational constant =
?? . ???? × ?? ?? -????
????
?? ????
-?? ) 
JEE Main 2025 (Online) 2nd April Evening Shift 
Ans: 3 
Solution: 
The kinetic energy (KE) of a satellite revolving around the Earth can be expressed with the following 
formula: 
KE =
1
2
?? ?? 2
=
1
2
?? ?? ?? ?? ?? =
?? ?? ?? ?? 2?? 
For this satellite, the radius ?? is the sum of the Earth's radius ?? ?? and the height h above the Earth's 
surface: 
?? = ?? ?? + h 
Given: 
Mass of the satellite, ?? = 1000 kg 
Mass of the Earth, ?? ?? = 6 × 10
24
 kg 
Radius of the Earth, ?? ?? = 6.4 × 10
6
 m 
Height of orbit above the Earth's surface, h = 270 km = 2.7 × 10
5
 m 
Gravitational constant, ?? = 6.67 × 10
-11
Nm
2
 kg
-2
 
Substitute these values into the equation: 
KE =
6.67 × 10
-11
× 6 × 10
24
× 1000
2( 6.4 × 10
6
+ 2.7 × 10
5
)
 
=
6.67 × 10
-11
× 6 × 10
24
× 1000
2 × 6.67 × 10
6
 
= 3 × 10
10
 J 
Hence, the kinetic energy of the satellite in its orbit is 3 × 10
10
 J. 
Q5: Three identical spheres of mass ?? , are placed at the vertices of an equilateral triangle of 
length ?? . When released, they interact only through gravitational force and collide after a 
time ?? = ?? seconds. If the sides of the triangle are increased to length ?? ?? and also the 
masses of the spheres are made 2 m , then they will collide after ____ seconds. 
JEE Main 2025 (Online) 3rd April Morning Shift 
Ans: 8 
Solution: 
Dimensional Analysis: 
?? ? ?? ?? ?? ?? ?? ?? 
Using dimensional analysis for gravitational interactions, we have: 
T ? M
x
?? y
?? z
 
Where ?? has dimensions [M
-1
 L
3
 T
-2
]. 
Solving for Exponents: 
T ? M
x-y
L
3?? +?? T
-2?? 
Equating dimensions, we solve: 
x - y = 0 ? x = y 
-2y = 1 ? y = -
1
2
, x = -
1
2
 
3y+ z = 0 ? z = -3y =
3
2
 
Time Proportionality: 
T ? m
-1/2
G
-1/2
a
3/2
 
Which simplifies to: 
T ? (
?? 3
?? )
1/2
 
Applying New Conditions: 
When the side length becomes 2?? and mass becomes 2?? : 
T = 4 × (
( 2?? )
3
2?? )
1/2
 
Simplifying: 
T = 4 × (
8?? 3
2?? )
1/2
= 4 × ( 4)
1/2
= 8 seconds  
Thus, the spheres will collide after 8 seconds under the new conditions. 
Q6: A small point of mass ?? is placed at a distance ?? ?? from the centre ' ?? '
 of a big uniform 
solid sphere of mass ?? and radius ?? . The gravitational force on ' ?? ' due to ?? is ?? ?? . A 
spherical part of radius ?? /?? is removed from the big sphere as shown in the figure and the 
gravitational force on ?? due to remaining part of ?? is found to be ?? ?? . The value of ratio 
?? ?? : ?? ?? is 
 
JEE Main 2025 (Online) 22nd January Morning Shift 
Options: 
A. 11: 10 
B. 12 : 11