All Exams  >   NEET  >   30-Day Revision Course for NEET  >   All Questions

All questions of Gravitation for NEET Exam

 Which is untrue about orbital velocity?
  • a)
    increases with the increase in height of satellite
  • b)
    depends on mass and radius of planet around which it revolves
  • c)
    it is independent of mass of satellite
  • d)
    decreases with an increase in radius of orbit
Correct answer is option 'A'. Can you explain this answer?

Om Desai answered
The untrue statement about orbital velocity is:

1. increases with the increase in height of satellite

Explanation: Orbital velocity is the speed at which an object revolves around a planet or other celestial body in a stable orbit. According to the equation for orbital velocity, v = √(GM/r+h), where G is the gravitational constant, M is the mass of the planet, and r is the radius of the orbit.

As the height of the satellite increases (meaning it gets far to the planet), its h increase , so reasulting in decrease in velocity .

The other statements are true:

2. depends on mass and radius of planet around which it revolves: As mentioned in the equation, orbital velocity depends on both the mass (M) of the planet and the radius (r) of the orbit.

3. it is independent of mass of satellite: The mass of the satellite does not appear in the equation for orbital velocity, so it does not affect the speed at which the satellite orbits the planet.

4. decreases with an increase in radius of orbit: From the equation, we can see that as the radius of the orbit (r) increases, the orbital velocity (v) decreases.

The escape velocity of a body depends upon mass as                   [AIEEE 2002]
  • a)
    m0
  • b)
    m1
  • c)
    m2
  • d)
    m3
Correct answer is option 'A'. Can you explain this answer?

Pooja Shah answered
We know that escape velocity, 
Where M is the mass of the planet and R is the radius of the planet.
Thus we can see that v does not depend upon the mass of the object.

A satellite which appears to be at a fixed position at a definite height to an observer is called:
  • a)
    Geostationary satellite and geosynchronous satellite
  • b)
    Polar satellite
  • c)
    Geostationary satellite
  • d)
    Geosynchronous satellite
Correct answer is option 'A'. Can you explain this answer?

Neha Joshi answered
As the relative velocity of the satellite with respect to the earth is zero, it appears stationary from the Earth surface and therefore it is called is geostationary satellite or geosynchronous satellite.

The mass of a spaceship is 1000kg. It is to be launched from the earth's surface out into free space. The value of g and R (radius of earth) are 10 m/s2 and 6400 Km respectively. The required energy for this work will be   
[AIEEE 2012]
  • a)
    6.4 x 1011 J
  • b)
    6.4 x 108 J
  • c)
     6.4 x 109J
  • d)
    6.4 x 1010J
Correct answer is option 'D'. Can you explain this answer?

Jithin Saini answered
Given,
Mass of spaceship (m) = 1000 kg
Acceleration due to gravity (g) = 10 m/s²
Radius of Earth (R) = 6400 km
We need to find out the required energy to launch the spaceship from Earth's surface to free space.

Potential Energy Calculation
When we lift an object against the gravitational force, the work done is stored as potential energy. The potential energy of an object at a height 'h' above the surface of the Earth is given by:
PE = mgh
where m is the mass of the object, g is the acceleration due to gravity and h is the height.

- Calculate the height attained by the spaceship
The spaceship is launched from the Earth's surface to free space. Therefore, the height attained by the spaceship is equal to the distance between the Earth's surface and the edge of the atmosphere.

- Calculate the radius of the atmosphere
The edge of the atmosphere is not clearly defined. However, the Kármán line is considered to be the boundary between the Earth's atmosphere and outer space. The Kármán line is located at a height of 100 km above the Earth's surface.

- Calculate the height of the spaceship
Height of the spaceship = Height of Kármán line - Radius of the Earth
Height of the spaceship = 100 km - 6400 km
Height of the spaceship = -6300 km (negative sign indicates that the spaceship is below the Earth's surface)

- Calculate the potential energy of the spaceship
PE = mgh
PE = 1000 kg x 10 m/s² x (-6300000 m)
PE = -6.3 x 10¹⁰ J (negative sign indicates that the potential energy is negative since the spaceship is below the Earth's surface)

Kinetic Energy Calculation
The kinetic energy of an object is given by:
KE = ½mv²
where m is the mass of the object and v is its velocity.

- Calculate the escape velocity of the spaceship
The escape velocity is the minimum velocity required to escape the gravitational field of the Earth. It is given by:
Vesc = √(2GM/R)
where G is the gravitational constant, M is the mass of the Earth and R is the distance between the spaceship and the center of the Earth.

- Calculate the distance between the spaceship and the center of the Earth
The distance between the spaceship and the center of the Earth is equal to the sum of the radius of the Earth and the height of the spaceship.

Distance between spaceship and center of Earth = Radius of Earth + Height of spaceship
Distance between spaceship and center of Earth = 6400 km - 6300 km
Distance between spaceship and center of Earth = 100 km

- Calculate the escape velocity of the spaceship
Vesc = √(2GM/R)
Vesc = √(2 x 6.67 x 10⁻¹¹ Nm²/kg² x 5.97 x 10²⁴ kg / 6.38 x 10⁶ m)
Vesc = 11.2 km/s

- Calculate the kinetic energy of the spaceship
KE = ½mv²
KE = ½ x 1000 kg x (11.2 km/s)²
KE = 6.2 x 10¹⁰ J

Total Energy Calculation
The total energy required

Kepler’s second law states that the straight line joining the planet to the sun sweeps out equal areas in equal time. The statement is equivalent to saying that:
  • a)
    longitudnal acceleration is zero
  • b)
    total acceleration is zero
  • c)
    transverse acceleration is zero
  • d)
    radial acceleration is zero
Correct answer is option 'C'. Can you explain this answer?

Suresh Reddy answered
According to the second law the orbital radius and angular velocity of the planet in the elliptical orbit will vary. The planet travels faster when closer to the Sun, then slower when farther from the Sun. Hence we can say that the transverse acceleration is zero while radial and longitudinal accelerations are not zero.

Different planets have different escape velocities because:
  • a)
    different planets have different atmosphere
  • b)
    they have different distances from sun
  • c)
    escape velocity depends on the body that has to escape the atmosphere of the planet
  • d)
    they have different masses and sizes
Correct answer is option 'D'. Can you explain this answer?

Nitin Nair answered
The formula for calculating the escape velocity from the surface of a celestial body (e.g. a planet) is:
where G is the universal gravitation constant, M is the planet’s mass and R is its radius.Different planets have different mass and radius - and therefore different escape velocity.

A satellite moves in a circular orbit around earth. The radius of this orbit is one half that of moon’s orbit. The satellite completes one revolution in:
  • a)
    (2) 3/2 lunar month
  • b)
    (2) 3 lunar month
  • c)
    (2) -3/2 lunar month
  • d)
    (2)1/2 lunar month
Correct answer is option 'C'. Can you explain this answer?

Suresh Reddy answered
We know that the time period of revolution for any object in orbit of radius r is 3
T = 2π (r3 / Gm) 1/2
Where m is the mass of earth.
As the ratio of radius of orbit of satellite to that of moon is 1:2
Hence there time period has a ratio of 1: 23/2

If gE and gM are the accelerations due to gravity on the surfaces of the earth and the moon respectively and if Millikan's oil drop experiment could be performed on the two surfaces, one will find the ratio  to be  
 [AIEEE 2007]
  • a)
    1
  • b)
    Zero
  • c)
  • d)
     
Correct answer is option 'A'. Can you explain this answer?

Geetika Shah answered
Since electronic charge (1.6 × 10–19C) is a universal constant. It does not depend on g. Therefore, electronic charge on the moon = electronic charge on the earth
Or
(electronic charge on the moon)/ (electronic charge on the earth) = 1

 If suddenly the gravitational force of attraction between earth and a satellite revolving around it becomes zero, then the satellite will   [AIEEE 2002]
  • a)
    Continue to move in its orbit with same velocity
  • b)
    Move tangentially to the original orbit with the same velocity
  • c)
    Become stationary in its orbit
  • d)
    Move towards the earth
Correct answer is option 'B'. Can you explain this answer?

Mira Sharma answered
When the gravitational force becomes zero that means the centripetal force required cannot be provided therefore the satellite will tend to move tangential to the original orbit  with the same velocity (that the satellite had at the instant before the gravitation force become 0).

Chapter doubts & questions for Gravitation - 30-Day Revision Course for NEET 2026 is part of NEET exam preparation. The chapters have been prepared according to the NEET exam syllabus. The Chapter doubts & questions, notes, tests & MCQs are made for NEET 2026 Exam. Find important definitions, questions, notes, meanings, examples, exercises, MCQs and online tests here.

Chapter doubts & questions of Gravitation - 30-Day Revision Course for NEET in English & Hindi are available as part of NEET exam. Download more important topics, notes, lectures and mock test series for NEET Exam by signing up for free.

Top Courses NEET

Related NEET Content