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Chapter: 11th Physics : Gravitation

Solved Example Problems for Physics: Gravitation

Numerical Problems, Conceptual Questions, Solved Example Problems for Physics: Gravitation, Explanation, formula, etc

Numerical Problems

 

1. An unknown planet orbits the Sun with distance twice the semi major axis distance of the Earth’s orbit. If the Earth’s time period is T1, what is the time period of this unknown planet?

Solution

The earth’s time period = T1

Semi major axis distance of earth = α1

Time period of unknown planet = T2

Semi major axis of unknown planet = 2α1,

Kepler’s III law


 T2 =2√2T1

Ans: T2 = 2 2T1

 

2. Assume that you are in another solar system and provided with the set of data given below consisting of the planets’ semi major axes and time periods. Can you infer the relation connecting semi major axis and time period?


Solution

For planet kurinji

Time period T12  α13

T1=2;

α1 = 8

α1 = 2.22 => 2.T12

Similarly for other planets

α2 = 2.32 => 2T22

α3 = 2.42 => 2T32

α4 = 2.52 => 2T42

α  2T2

Ans: a  2T2

 

3. If the masses and mutual distance between the two objects are doubled, what is the change in the gravitational force between them?

Answer


Now the masses and distance between two objects is doubled.


F2 = F1

No change in gravitational force when masses and distance are doubled.

Ans: No change

 

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


Ans: V = −9Gm/r

 

5. If the ratio of the orbital distance of two planets d1/d2 =2, what is the ratio of gravitational field experienced by these two planets?


Ans: E2 = 4 E1

 

6. The Moon Io orbits Jupiter once in 1.769 days. The orbital radius of the Moon Io is 421700 km. Calculate the mass of Jupiter?

Solution

Time period of moon Io = 1.769 days

= 1.769 × 24 × 60 × 60s.

Orbital radius of the moon (r) = 4.217 × 108m.


M = 1.898 × 1027kg

Ans: 1.898 ×1027 kg

 

7. If the angular momentum of a planet is given by  = 5 t 2iˆ 6tjˆ +3kˆ. What is the torque experienced by the planet? Will the torque be in the same direction as that of the angular momentum?



 

8. Four particles, each of mass M and equidistant from each other, move along a circle of radius R under the action of their mutual gravitational attraction. Calculate the speed of each particle



 

9. Suppose unknowingly you wrote the universal gravitational constant value as G = 6.67x1011 instead of the correct value G = 6.67x10-11, what is the acceleration due to gravity g' for this incorrect G? According to this new acceleration due to gravity, what will be your weight W'?

Solution:

Mass of the earth M = 6.024 × 1024 kg

Radius of the earth R = 6.4 × 106m

Gravitational constant G' (Incorrect) = 6.67 × 1011

Gravitational constant G (Correct) = 6.67 × 1011

Acceleration due to gravity g' = ?


Ans: g' =1022 g, W' = 1022W

 

10. Calculate the gravitational field at point O due to three masses m1,m2 and m3 whose positions are given by the following figure. If the masses m1 and m2 are equal what is the change in gravitational field at the point O?



 

11. What is the gravitational potential energy of the Earth and Sun? The Earth to Sun distance is around 150 million km. The mass of the Earth is 5.9 × 1024 kg and mass of the Sun is 1.9 × 1030 kg.

Solution:

The distance between sun & Earth R= 150 million km = 150 × 1011m.

The mass of the sun Ms = 1.9 × 1030 kg

The mass of the Earth Me = 5.9 × 1024 kg

The gravitational potential energy U = ?


U = -4.985 × 1032 J

P.E U = -49.84 × 1031 J

Ans: V = -49.84 × 1032 Joule

 

12. Earth revolves around the Sun at 30 km s−1. Calculate the kinetic energy of the Earth. In the previous example you calculated the potential energy of the Earth. What is the total energy of the Earth in that case? Is the total energy positive? Give reasons.

Solution:

The velocity of earth revolves around the sun = v = 30kms (or) 30 × 103 ms-1

K.E of the earth = 1/2 mv2 = ?

Mass of the earth = 5.9 × 1024 kg

K.E = 1/2 × 5.9 × 1024 × 30 × 30 × 1O+3 × 10+3

= 1/2 × 5.9 × 1024 × 900 × 106

= 2655 × 1030 (or) 26.55 × 1032J

T.E = P.E + K.E

= - 49.84 × 1032 + 26.55 × 1032

E = - 23.29 × 1032 J 

Ans: K.E = 26.5 × 1032 J

E = −23.29 × 1032 J

( - ) ve implies that Earth is bounded with Sun

 

13. An object is thrown from Earth in such a way that it reaches a point at infinity with non-zero kinetic energy

  with what velocity should the object be thrown from Earth?



 

14. Suppose we go 200 km above and below the surface of the Earth, what are the g values at these two points? In which case, is the value of g small? 


Ans: gdown = 0.96 g

gup = 0.94 g

 

15.  Calculate the change in g value in your district of Tamil nadu. (Hint: Get the latitude of your district of Tamil nadu from the Google). What is the difference in g values at Chennai and Kanyakumari?

Solution:

Chennai

g'c = g - ω2 R cos2λ

 = 9.8 - (3.4 × 10-2)

= 9.7677 ms-2

g'K =9.8 - (3.4x 10-2) × (cos0.1396)2 = 9.748 ms-2

Difference in g value Δg = 0.0331 ms-2

ω2R = (2 × (3.14/86400) )2 × 6400 × 103

λ =13' (or) 0.2268 rad 

Ans: gchennai = 9.767 m s-2

gKanyakumari = 9.798 m s-2

Δg = 0.031 m s-2


Conceptual Questions

 

1. In the following, what are the quantities which that are conserved?

a) Linear momentum of planet

b) Angular momentum of planet

c) Total energy of planet

d) Potential energy of a planet

Answer: (b & d) Angular momentum of planet, Potential energy of a planet]

 

2. The work done by Sun on Earth in one year will be

a) Zero

b) Non zero

c) positive

d) negative

[ Answer: (a) zero]

 

3. The work done by Sun on Earth at any finite interval of time is

a) positive, negative or zero

b) Strictly positive

c) Strictly negative

d) It is always zero

[Answer: (d) It is always zero]

 

4. If a comet suddenly hits the Moon and imparts energy which is more than the total energy of the Moon, what will happen?

Answer

If a comet hits the moon with large mass and with large velocity may destroy the moon completely or its impact makes the moon, go out of the orbit.

 

5.  If the Earth’s pull on the Moon suddenly disappears, what will happen to the Moon?

Answer

If the gravitational force suddenly disappears, moon will stop revolving around the earth and it will move in a direction tangential to its original orbit with a speed with which it was revolving around the earth.

 

6.  If the Earth has no tilt, what happens to the seasons of the Earth?

Answer

If the Earth has no tilt, there will be no seasons like now and the duration of day and night \vi! be equal throughout the year.

 

7.  A student was asked a question ‘why are there summer and winter for us? He replied as ‘since Earth is orbiting in an elliptical orbit, when the Earth is very far away from the Sun(aphelion) there will be winter, when the Earth is nearer to the Sun(perihelion) there will be winter’. Is this answer correct? If not, what is the correct explanation for the occurrence of summer and winter?

Answer

No, The seasons in the Earth arise due to tit; rotation of Earth around the Sun with 23.5° tilt Due to this 23.5° tilt, when the northern parto Earth is farther to the Sun, the southern part: nearer to the Sun. So when it is summer in tb northern hemisphere, the southern hemisphet experience winter.

 

8.  The following photographs are taken from the recent lunar eclipse which occurred on January 31, 2018. Is it possible to prove that Earth is a sphere from these photographs?

 

Answer

No. The moon goes around the earth in an elliptical orbit. This means its distance from us varies periodically as it goes around us.


EXAMPLE 6.1

Consider two point masses mand m2 which are separated by a distance of 10 meter as shown in the following figure. Calculate the force of attraction between them and draw the directions of forces on each of them. Take m1= 1 kg and m2 = 2 kg


Solution

The force of attraction is given by


From the figure, r =10 m.

First, we can calculate the magnitude of the force


It is to be noted that this force is very small. This is the reason we do not feel the gravitational force of attraction between each other. The small value of G plays a very crucial role in deciding the strength of the force.

The force of attraction () experienced by the mass m2 due to m1 is in the negative ‘y’ direction ie., rˆ =−jˆ . According to Newton’s third law, the mass m2 also exerts equal and opposite force on m1. So the force of attraction () experienced by m1 due to m2 is in the direction of positive ‘y’ axis ie., rˆ = jˆ .


The direction of the force is shown in the figure,


Gravitational force of attraction between m1 and m2

=  which confirms Newton’s third law.

 

EXAMPLE 6.2

Moon and an apple are accelerated by the same gravitational force due to Earth. Compare the acceleration of the two.

The gravitational force experienced by the apple due to Earth


Here M– Mass of the apple, ME– Mass of the Earth and R – Radius of the Earth.

Equating the above equation with Newton’s second law,


Simplifying the above equation we get,


Here aA is the acceleration of apple that is equal to ‘g’.

Similarly the force experienced by Moon due to Earth is given by


Here Rm- distance of the Moon from the Earth, Mm – Mass of the Moon

The acceleration experienced by the Moon is given by


The ratio between the apple’s acceleration to Moon’s acceleration is given by


From the Hipparchrus measurement, the distance to the Moon is 60 times that of Earth radius. Rm = 60R.


The apple’s acceleration is 3600 times the acceleration of the Moon.

The same result was obtained by Newton using his gravitational formula. The apple’s acceleration is measured easily and it is 9.8 m s−2 . Moon orbits the Earth once in 27.3 days and by using the centripetal acceleration formula, (Refer unit 3).


which is exactly what he got through his law of gravitation.


EXAMPLE 6.3

 

a) Two particles of masses m1 and m2 are placed along the x and y axes respectively at a distance ‘a’ from the origin. Calculate the gravitational field at a point P shown in figure below.


Solution

Gravitational field due to m1 at a point P is given by,


Gravitational field due to m2 at the point p is given by,


The direction of the total gravitational field is determined by the relative value of m1 and m2.

When m1 = m2 = m


(iˆ + jˆ = jˆ + iˆ as vectors obeys commutation law).


total points  towards  the  origin  of  the co-ordinate system and the magnitude of total is Gm/a2

 

EXAMPLE 6.4

Qualitatively indicate the gravitational field of Sun on Mercury, Earth, and Jupiter shown in figure.

Since the gravitational field decreases as distance increases, Jupiter experiences a weak gravitational field due to the Sun. Since Mercury is the nearest to the Sun, it experiences the strongest gravitational field.


EXAMPLE 6.5

Water falls from the top of a hill to the ground. Why?

This is because the top of the hill is a point of higher gravitational potential than the surface of the Earth i.e.Vhill > Vground


The motion of particles can be analyzed more easily using scalars like U ( r )or V(r)  than vector quantities like  or . In modern theories of physics, the concept of potential plays a vital role.

 

EXAMPLE 6.6

Consider four masses m1, m2, m3, and m4 arranged on the circumference of a circle as shown in figure below


Calculate

a) The gravitational potential energy of the system of 4 masses shown in figure.

b) The gravitational potential at the point O due to all the 4 masses.

Solution

The gravitational potential energy U( r ) can be calculated by finding the sum of gravitational potential energy of each pair of particles.


If all the masses are equal, then m1 = m2 = m3 = m4 = M



The gravitational potential V(r) at a point O is equal to the sum of the gravitational potentials due to individual mass. Since potential is a scalar, the net potential at point O is the algebraic sum of potentials due to each mass.


EXAMPLE 6.7

1. Calculate the value of g in the following two cases: 

a) If a mango of mass ½ kg falls from a tree from a height of 15 meters, what is the acceleration due to gravity when it begins to fall? 

Solution

 

b) Consider a satellite orbiting the Earth in a circular orbit of radius 1600 km above the surface of the Earth. What is the acceleration experienced by the satellite due to Earth’s gravitational force?

Solution


The above two examples show that the acceleration due to gravity is a constant near the surface of the Earth.

 

EXAMPLE 6.8

Find out the value of g in your school laboratory?

Solution

Calculate the latitude of the city or village where the school is located. The information is available in Google search. For example, the latitude of Chennai is approximately 13 degree.

′ = g −ω 2 R cos2 λ

Here ω2R = (2x3.14/86400)2 x (6400x103) = 3.4x10−2 m s−2.

It is to be noted that the value of λ should be in radian and not in degree. 13 degree is equivalent to 0.2268 rad.

g′ = 9.8  ( 3.4 × 102 × ( cos 0.2268)2

g = 9.7677 m s−2


EXAMPLE 6.9

Moon is the natural satellite of Earth and it takes 27 days to go once around its orbit. Calculate the distance of the Moon from the surface of the Earth assuming the orbit of the Moon as circular.

Solution

We can use Kepler’s third law,


Here h is the distance of the Moon from the surface of the Earth. Here,


By substituting these values, the distance to the Moon from the surface of the Earth is calculated to be 3.77 ×105 km.


EXAMPLE 6.10

Calculate the energy of the (i) Moon orbiting the Earth and (ii) Earth orbiting the Sun.

Solution

Assuming the orbit of the Moon to be circular, the energy of Moon is given by,


where  ME is the mass  of Earth 6.02 ×1024  kg; Mm is the mass of Moon 7.35 ×1022 kg; and R  is the distance between the Moon and the center of the Earth 3.84 ×105 km


The negative energy implies that the Moon is bound to the Earth.

Same method can be used to prove that the energy of the Earth is also negative.


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11th Physics : Gravitation : Solved Example Problems for Physics: Gravitation |


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