Newton’s law of gravitation applies to

All bodies irrespective of their size

Two objects of different masses falling freely near the surface of moon would:

Have same velocities at any instant

Undergo a change in their inertia

Experience forces of same magnitude

JEE Main Previous Year Solved Questions on Gravitation

Gravitation is a natural phenomenon. It is the force of attraction between any two objects with mass or energy. The gravitational force brings all the objects in the universe like planets, stars and galaxies together. The tides in the ocean are caused due to the gravitational force of attraction of the moon. But in most cases, the force is too weak to be observed due to the very large distance of separation. The effect of gravity reduces as the objects move away from each other. The weight of physical objects on earth is given as the product of its mass and the acceleration due to the gravity of the earth. Read More.

This force of attraction was first observed by Sir Isaac Newton and was presented as Newton’s law of gravitation in the year 1680. However, gravitation can generally exist in two main instances.

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JEE Main Previous Year Solved Questions on Gravitation

Q1: If the distance between the earth and the sun were half its present value, the number of days in a year would have been

(a) 64.5

(b) 129

(d) 730

Solution: From Kepler’s law, T²∝ R³

Therefore,

\(\begin{array}{l}\left ( \frac{T_{2}}{T_{1}} \right )^{2} = \left ( \frac{R_{2}}{R_{1}} \right )^{3}\end{array} \)

T₁= 365 days , R₁ =R , R₂ = R/2

\(\begin{array}{l}\left ( \frac{T_{2}}{365} \right )^{2} = \left ( \frac{R/2}{R} \right )^{3}\end{array} \)

T₂² = (365)²/8

T₂² = 16,653

T₂ = 129 days

Answer: (b)129 days

Q2: The mass of a spaceship is 1000 kg. It is to be launched from the earth’s surface out into free space. The value of ‘g’ and ‘R’ (radius of the earth) is 10 m/s² and 6400 km respectively. The required energy for this work will be

(a) 6.4 x 10¹⁰Joules

(b) 6.4 x 10¹¹Joules

(d) 6.4 x 10⁹Joules

Solution

The energy required is given by = GMm/R

= gR² x m/R (∵ g = GM/R²)

= mgR

= 1000 x 10 x 6400 x 10³

= 64 x 10⁹ J

= 6.4 x 10¹⁰ J

Answer: (a) 6.4 x 10¹⁰Joules

Q3: An artificial satellite moving in a circular orbit around the earth has a total (K.E. + P.E.) energy E₀. Its potential energy is

(a) – E₀

_(b)1.5 E₀

(d) E₀

Solution

Total Energy, E₀ = – GMm/2r

Potential Energy, U = -GMm/r = 2E₀

Answer: (c) 2 E₀

Q4: The height at which the acceleration due to gravity becomes g/9 (where g = the acceleration due to gravity on the surface of the earth) in terms of R, the radius of the earth, is

(a) R/2

(b) R/3

(d) 3R

Solution

Acceleration due to gravity at a height “h” is given by

g’ = g (R/R+h)²

Here,

g is the acceleration due to gravity on the surface

R is the radius of the earth

As g’ is given as g/9, we get

g/9 = g(R/R+h)²

⅓ = R/(R+h)

h=2R

Answer: (c) 2R

Q5: A simple pendulum has a time period T₁ when on the earth’s surface, and T₂ when taken to a height R above the earth’s surface, where R is the radius of the earth. The value of T₂/T₁ is

(a) 1

(b) 3

(d) 2

Solution

The time period of a simple pendulum

\(\begin{array}{l}=2\pi \sqrt{\frac{l}{g}}\end{array} \)

On the surface of earth g₁ = GM/R²

At a height R above the earth g₂= GM/(2R)²

(g₁/g₂) = (4/1)

T = 2π√(l/g)

The time period on the surface of the earth T₁

\(\begin{array}{l}=2\pi \sqrt{\frac{l(R)^{2}}{GM}}\end{array} \)

The time period on the surface of the earth T₂

\(\begin{array}{l}=2\pi \sqrt{\frac{l(2R)^{2}}{GM}}\end{array} \)

T₂/T₁ = 2

Answer: (d) 2

Q6: A geostationary satellite orbits around the earth in a circular orbit of radius 36,000km. Then, the time period of a spy satellite orbiting a few hundred km above the earth’s surface (R =6,400km) will approximately be

(a) (l/2)hr

(b) 1 hr

(d) 4 hr

Solution

According to Kepler’s law, T²∝ R³

Therefore,

\(\begin{array}{l}\left ( \frac{T_{2}}{T_{1}} \right )^{2} = \left ( \frac{R_{2}}{R_{1}} \right )^{3}\end{array} \)

For a spy satellite time period is given by T₁

R₁ = 6400 km

For a geostationary satellite, T₂ = 24hour

R₂ = 36,000 km

\(\begin{array}{l}\left ( \frac{24}{T_{1}} \right )^{2} = \left ( \frac{36000}{6400} \right )^{3}\end{array} \)

(24/T₁)²= 178

(24/T₁) = 13.34

T₁ = 24/13.34 ≈ 2hr

Answer: (c) 2 hr

Q7: Two bodies of masses m and 4m are placed at a distance r. The gravitational potential at a point on the line joining them where the gravitational field is zero is

(a) -6Gm/r

(b) -9Gm/r

(d) -4Gm/r

Solution

P is the point where the field is zero, and a unit mass is placed at P.

Applying Newton’s law of gravitation,

(Gm x 1)/x² = (G4m x 1)/(r – x)²

x²/(r – x)²= ¼

x/(r – x)= ½

2x = r – x

x = r/3

Potential at the point P, V = – Gm/x – G(4m)/(r-x)

\(\begin{array}{l}V = -\frac{Gm}{r/3} – \frac{G(4m)}{(r-r/3)}\end{array} \)

V = -9Gm/r

Answer: (b) -9Gm/r

Q8: A Binary star system consists of two stars A and B which have time periods T_A and T_B, radii R_A and R_B and masses M_A and M_B. Then

(a) T_A ＞T_B then R_A ＞R_B

(b) T_A ＞T_B then M_A ＞M_B

(d) T_A = T_B

Solution

Angular velocity of binary stars are the equal ω_A = ω_B

2π/T_A = 2π /T_B

⇒ T_A = T_B

Answer: (d) T_A = T_B

Q9: 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

\(\begin{array}{l}(a)\ \sqrt{Gm/R}\end{array} \)

\(\begin{array}{l}(b)\ \sqrt{Gm/4R}\end{array} \)

\(\begin{array}{l}(c)\ \sqrt{Gm/3R}\end{array} \)

\(\begin{array}{l}(d)\ \sqrt{Gm/2R}\end{array} \)

Solution

Gm²/4R² = mV²/R

\(\begin{array}{l}V= \sqrt{Gm/4R}\end{array} \)

Answer: (b)

Q10: A satellite is moving with a constant speed ‘V’ in a circular orbit about the earth. An object of mass “m” is ejected from the satellite such that it just escapes from the gravitational pull of the earth. At the time of its ejection, the kinetic energy of the object is

(a) 1/2 mV²

(b) mV²

(d) 2 mV²

Solution

Kinetic energy at the time of ejection = ½ mv_e²

V_e is the escape velocity = √2 x orbital velocity

\(\begin{array}{l}=\sqrt{2}v\end{array} \)

Kinetic energy at the time of ejection

\(\begin{array}{l}=\frac{1}{2}m(\sqrt{2}v)^2\end{array} \)

= mv²

Answer: (b) mv²

Q11: Three particles, each of mass m, are situated at the vertices of an equilateral triangle of side length a. The only forces acting on the particles are their mutual gravitational forces. lt is desired that each particle moves in a circle while maintaining the original mutual separation a. Find the initial velocity that should be given to each particle and also the time period of the circular motion. (F = Gm₁m₂/r²)

Solution:

ABC is an equilateral triangle of sides a.

Each particle moves in a circle of radius r

JEE Main Previous Year Questions on Gravitation

AD² = a² -a²/4, r = ⅔AD

\(\begin{array}{l}r=\frac{2}{3}\times \sqrt{a^{2}-a^{2}/4}\end{array} \)

\(\begin{array}{l}r= a\sqrt{3}…..(1)\end{array} \)

To find v

JEE Main Past Year Solved Questions on Gravitation

Let v= Initial velocity given to each particle.

For circular motion, centripetal force should be provided. It is provided by the gravitational force between two masses. Let F denote this force.

F = Gm²/a²——-(2)

Resultant force

\(\begin{array}{l}=\sqrt{F^{2} + F^{2} +2F^{2}cos60^{0}}\end{array} \)

Resultant force

\(\begin{array}{l}=\sqrt{3}F\end{array} \)

Resultant force = Centripetal force

\(\begin{array}{l}\sqrt{3}F\end{array} \)

=mV²/r

\(\begin{array}{l}V^{2}=\frac{\sqrt{3}Fr}{m}=\frac{\sqrt{3}}{m}\times \left ( \frac{Gm^{2}}{a^{2}} \right )\left ( \frac{a}{\sqrt{3}} \right ) = Gm/a\end{array} \)

\(\begin{array}{l}V=\sqrt{\frac{Gm}{a}}\end{array} \)

2. To find time period of circular motion(T)

T = 2πr/V

\(\begin{array}{l}=2\pi (a/\sqrt{3})\times \sqrt{a/Gm}=2\pi \sqrt{a^{3}/3Gm}\end{array} \)

Therefore,

\(\begin{array}{l}T=2\pi \sqrt{a^{3}/3Gm}\end{array} \)

Answer:

\(\begin{array}{l}V=\sqrt{\frac{Gm}{a}}\end{array} \)

and
\(\begin{array}{l}T=2\pi \sqrt{a^{3}/3Gm}\end{array} \)

Q12: What is the minimum energy required to launch a satellite of mass m from the surface of a planet of mass M and radius R in a circular orbit at an altitude of 2R?

(a) 5GmM/6R

(b) 2GmM/3R

(d) GmM/3R

Solution

Energy of the satellite on the surface of the Earth E₁ = -GMm/R

Energy at a distance 2R is given by E₂ = -GMm/3R + ½mv_o²

E₂ = -GMm/3R + ½m[GM/3R]

E₂ = -GmM/6R

E₂ -E₁ = (-GmM/6R) – ( -GMm/R ) = 5GmM/6R

Answer: (a) 5GmM/6R

Q13: Suppose that the angular velocity of rotation of earth is increased. Then,as a consequence

(a) There will be no change in weight anywhere on the earth

(b) Weight of the object, everywhere on the earth, will increase

(d) Weight of the object, everywhere on the earth, will decrease.

Solution

The effect of rotation of earth on acceleration due to gravity is given by g’ = g – ω²Rcos2Φ

Where Φ is latitude. There will be no change in gravity at poles as Φ = 90⁰, at all points as ω increases g’ will decrease.

Answer: (c) Except at poles, weight of the object on the earth will decrease

Q14: A particle is moving with a uniform speed in a circular orbit of radius R in a central force inversely proportional to the n^th power of R. If the period of rotation of the particle is T, then

(a) T ∝ R^3/2

(b) T ∝ R^(n/2)+1

(d) T ∝ R^n/2

Solution

According to the question, central force is given by

F_c ∝ 1/Rⁿ

F_c =k (1/Rⁿ)

mω²R = k(1/Rⁿ)

\(\begin{array}{l}m\frac{(2\pi )^{2}}{T^{2}}=k\frac{1}{R^{n+1}}\end{array} \)

Or T² ∝ Rⁿ⁺¹

T ∝ R^(n+1)/2

Answer: (c) T ∝ R^(n+1)/2

Also Read:

Gravitation JEE Advanced Previous Year Questions With Solutions

JEE Main Gravitation Previous Year Questions with Solutions

JEE Main Previous Year Solved Questions on Gravitation

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