Gravitation is a natural phenomenon. It is the force of attraction between any two objects with mass or energy. Gravitational force that 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 of the 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 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

(c) 182.5

(d) 730

**Solution: **From Kepler’s law, T^{2}∝ R^{3}

Therefore,

T_{1 }= 365 days , R_{1} =R , R_{2} = R/2

T_{2}^{2} = (365)^{2}/8

T_{2}^{2} = 16,653

T_{2} = 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/s ^{2} and 6400 km respectively. The required energy for this work will be **

(a) 6.4 x 10^{10 }Joules

(b) 6.4 x 10^{11 }Joules

(c) 6.4 x 10^{8 }Joules

(d) 6.4 x 10^{9 }Joules

**Solution**

The energy required is given by = GMm/R

= gR^{2} x m/R (∵ g = GM/R^{2})

= mgR

= 1000 x 10 x 6400 x 10^{3}

= 64 x 10^{9} J

= 6.4 x 10^{10} J

**Answer: (a) 6.4 x 10 ^{10 }Joules**

**Q3: An artificial satellite moving in a circular orbit around the earth has a total (K.E. + P.E.) energy E _{0}. Its potential energy is **

(a) – E_{0}

_{(b) }1.5 E_{0}

(c) 2 E_{0}

(d) E_{0}

**Solution**

Total Energy, E_{0} = – GMm/2r

Potential Energy, U = -GMm/r = 2E_{0}

**Answer: ** **(c) 2 E _{0} **

**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

(c) 2R

(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

(c) 4

(d) 2

**Solution**

The time period of a simple pendulum =

On the surface of earth g = GM/R^{2}

At a height R above the earth g = GM/(2R)^{2}

The time period on the surface of the earth T1 =

The time period on the surface of the earth T2 =

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

(c) 2 hr

(d) 4 hr

**Solution**

According to Kepler’s law, T^{2}∝ R^{3}

Therefore,

For a spy satellite time period is given by T1

R1 = 6400 km

For a geostationary satellite, T2 = 24hour

R2 = 36,000 km

(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

(c) Zero

(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)/x^{2} = (G4m x 1)/(r – x)^{2}

x^{2}/(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)

V = -9Gm/r

**Answer: (b) -9Gm/r**

**Q8: A Binary star system consists of two stars A and B which have time periods TA and TB, radii RA and RB and masses MA and MB. Then**

(a) T_{A} ＞T_{B} then R_{A} ＞R_{B}

(b) T_{A} ＞T_{B} then M_{A} ＞M_{B }

(c) (T_{A}/T_{B})^{2} = (R_{A}/R_{B})^{2}

(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**

(a)

(b)

(c)

(d)

**Solution**

Gm^{2}/4R^{2} = mV^{2}/R

**Answer: (b) $\sqrt{Gm/4R}$**

**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 mV^{2 }

(b) mV^{2}

(c) 3/2 mV^{2}

(d)2 mV^{2}

**Solution**

Kinetic energy at the time of ejection = ½ mv_{e}^{2 }

V_{e } is the escape velocity =

V_{e } is the escape velocity =

Kinetic energy at the time of ejection = ½ m( ^{2}

= mv^{2}

**Answer: (b) mv ^{2}**

**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 = Gm _{1}m_{2}/r^{2})**

**Solution: **

ABC is a equilateral triangle of side a

Each particle moves in a circle of radius r

AD^{2} = a^{2} -a^{2}/4, r = ⅔AD

- To find v

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 = Gm^{2}/a^{2}——-(2)

Resultant force =

Resultant force =

Resultant force = Centripetal force

^{2}/r

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

T = 2πr/V =

therefore ,

**Answer: $V=\sqrt{\frac{Gm}{a}}$ and $T=2\pi \sqrt{a^{3}/3Gm}$**

**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

(c) GmM/2R

(d) GmM/3R

**Solution**

Energy of the satellite on the surface of the Earth E_{1} = -GMm/R

Energy at a distance 2R is given by E_{2} = -GMm/3R + ½mv_{o}^{2}

E_{2} = -GMm/3R + ½m[GM/3R]

E_{2} = -GmM/6R

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

**Answer: (a) 5GmM/6R**

**Q13: Two satellites S1 and S2 revolve around a planet in coplanar circular orbits in the same sense. Their periods of revolution are 1 hour and 8 hours, respectively. The radius of the orbit of S1 is 10 ^{4} km. When S2 is closest to S1, find the angular speed of S2 as actually observed by an astronaut in S1**

**Solution:**

Angular velocity ω = 2π/T

Here T = 1 hour for S_{1},ω1=2π rad/hr

Similarly, for S_{2},ω2 = 2π

Given that they are rotating in the same sense. So, relative angular velocity = 2π – 2π/8= 7π/8

**Answer: Angular speed = 7π/8**

**Q14: 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

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

(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 – ω^{2}Rcos2Φ

Where Φ is latitude. There will be no change in gravity at poles as Φ = 90^{0}, at all points as ω increases g’ will decrease.

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

**Q15: 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}

(c) T ∝ R^{(n+1)/2}

(d) T ∝ R^{n/2}

**Solution**

According to the question, central force is given by

F_{c} ∝ 1/R^{n}

F_{c} =k (1/R^{n})

mω^{2}R = k(1/R^{n})

Or T^{2} ∝ R^{n+1}

T ∝ R^{(n+1)/2}

**Answer: (c) T ∝ R ^{(n+1)/2}**