# Important Questions Class 11 Physics Chapter 8

## Important Questions Class 11 Physics Chapter 8

### Important Questions for CBSE Class 11 Physics Chapter 8 – Gravitation

Chapter 8 of Class 11 Physics is a very crucial chapter for the students. This chapter’s Important Question will aid in the analysis and evaluation of the students.Class 11 Physics Chapter 8 Important Questions will help you take the first step in this self-evaluation. These questions are designed to help in comprehending the types of questions that will appear on the examination. One can assess themselves and get a firm grasp of the chapter by solving these questions. This increases your confidence and helps you manage your time for the exam.

Some of the main topics discussed in Class 11 Physics Chapter 8 Gravitation are as follows.

1. Kepler’s laws
2. Universal law of gravitation
3. The gravitation constant
4. Acceleration due to:
1. Gravity of the Earth
2. Gravity below and above the Earth’s surface
5. Gravitational potential energy
6. Escape speed
7. Earth satellites
8. Energy of an orbiting satellite
9. Geostationary and polar satellites
10. Weightlessness

Extramarks offers Important Questions for Class 11 Physics Chapter 8 that cover these topics as well as other important aspects for a well-rounded preparation.

### Important Questions for Class 11 Physics Chapter 8 – Gravitation

A few of the important questions are discussed below. You can click on the link to access the complete set of questions for Chapter 8 Class 11 Physics Important Questions.

Q1. Explain gravitation.

Ans: Any two bodies will be attracted to one another by the force of gravity, also known as gravity. There is an attraction between everything in the cosmos, however most of the time the force is too faint to be seen due to the extreme distance between the objects.

Q2. Name two factors that influence whether or not a planet has an atmosphere.

Ans: The existence of an atmosphere on a planet is determined by two factors.

1.  The temperature on the planet’s surface
2. The presence of gravitational acceleration on the planet’s surface

Q3. At the centre of the Earth, a body is weightless. Why?

Ans: G = 0 in the centre of the Earth.

It is determined by:

w = mg

w = 0

The body is therefore weightless at the centre of the Earth.

### 3 Marks Answers and Questions

Q1. How far from Earth’s surface does the value of g decrease to 4 of its value on Earth’s surface, given that the Earth’s radius is 6400 kilometres?

Ans: The gravity at height h is determined by

gh = gRR+h2… (1)

Given.

gh = 4 … (2)

Radius of the Earth, R = 6400 km

From (1) and (2),

4g100=gRR+h2

4100=RR+h2

210=RR+h

2R+2h = 10R

2h = 8R

h = 4R = 4×6400

h = 25,600 km

Therefore, when the value of g is lowered to 4 of its value on the surface of the Earth, 25600 km is the height from that point.

Q2. Find expressions for an artificial satellite.

• Potential Energy

Ans: Potential energy of a satellite is

U=rF dx

Where, F is the force,

U=rGMmx2dx

Where,

M is the mass of the Earth,

m  is the mass of the satellite,

x is the distance of the satellite from the Earth’s surface,

U=GMmr1x2dx

U=GMm 1xr

U=GMm 1r+1

U= –GMmr

• Kinetic Energy

Ans: Kinetic energy is given as

K.E. = 12mv2

Where,

m is the mass

v is the velocity

However, for artificial satellites, velocity is given as

v=GMr

Where,

G is the gravitational constant,

M is the mass of the Earth, and

r is the radius of the Earth

K.E.=12m GMr

K.E.=K=GMm2r

• Total Energy

Ans: Total energy of a satellite is given as

TE = U+K

TE= –GMmr+GMm2r

TE= –GMm2r

Q3. Imagine if there was a planet that orbited the sun twice as quickly as the Earth. What would the size of its orbit be in relation to the Earth’s?

Ans: Time taken by the Earth to complete on revolution, Te = 1 year

Orbital radius of the Earth in its orbit, Re=1AU

Time taken by the planet to complete one revolution, Tp = 12 Te = 12 year

Orbital radius of the planet = Rp

From Kepler’s third law of planetary motion,

RpRe3= TpTe2

RpRe=TpTe23

RpRe=12123

RpRe=0.523

RpRe=0.63

As a result, the planet’s orbital radius will be 0.63 times smaller than the Earth’s.

### 4 Marks Answers and Questions

Q1. Assume that there are 2.5 1011 stars in our galaxy, each with a solar mass. How long will it take a star 50,000 light years from the galactic centre to make one revolution? Assume that the Milky Way is 105ly in diameter.

Ans: Mass of the Milky Way galaxy, M = 2.5 × 1011 solar mass

Solar Mass = Mass of the Sun  = 2.0 × 1036 kg

Mass of our galaxy, M = 2.5 × 1011 × 2 × 1035

M = 5 × 1041 kg

Diameter of the Milky Way, d = 105 ly

Radius of the Milky Way, r = 5 × 104 ly

1 ly = 9.46 × 1015m

r = 5 × 104 × 9.46 × 1015

r = 4.73 × 1020m

Since a star rotates around the galactic centre of the Milky Way, its time period is given by

T=42r3GM12=3.1424.73310606.67×10-11×5×1041

T=125.27×103012

T = 1.12 × 1016s

As we know,

1 year = 365 × 24 × 60 × 60s

We get,

1s = 1365246060years

1.12 × 1016s = 1.121016365246060

∴1.121016s=3.55108years

Thus, the star will take 3.55 × 108 years to complete one revolution.

Q2. A body weighs 63N while it is on the Earth’s surface. What is the magnitude of the Earth’s gravitational pull on it at a height equal to one-half the radius of the Earth?

Ans: Weight of the body, W = 63N

Acceleration due to gravity at height h from the Earth’s surface

Re = Radius of the Earth

For h=Re2, gravity at h is given by

gh= g1+hRe2

gh= g1+122

gh= 49g

Weight of a body of mass m at height h is given as”

W’ = mgh

W’ = 49mg

W’ = 49W

W’ = 49×63

W’ = 28N

Therefore, at a height equal to half the Earth’s radius, the gravitational force exerted on the body by the Earth is 28N.

Q3. A projectile can escape from the Earth’s surface at a speed of 11.2 km/s. With three times this speed, a body is thrust out. What is the speed of a body travelling far from Earth? Ignore the sun’s and other planets’ presence.

Ans: The escape velocity of the projectile from the Earth, vesc=11.2kms-1

Projectile velocity of the projectile, vp=3vesc

Mass of the projectile = m

Projectile’s velocity far from the Earth = vf

Projectile’s total energy on the Earth = 12mvp212mvesc2

The projectile’s gravitational potential energy far from the Earth is 0.

Total energy of the projectile far from the Earth =  12mvf2

From the law of energy conservation, we have

12mvp212mvesc2= 12mvf2

We get,

vf=vp2vesc2

vf=3vesc2vesc2

vf=8vesc

vf=8×11.2

vf=31.68kms-1

Therefore, the speed of the body far from the Earth’s surface is 31.68kms-1

### 5 Marks Answers and Questions

Q1. One of Jupiter’s satellites, IO, has an orbital period of 1.769 days and an orbital radius of 4.22 108 metres. Prove that Jupiter’s mass is around one thousandth that of the sun.

Ans: Given,

Orbital period of IO=TIO = 1.769 days = 1.769×24×60×60s

Satellite IO is revolving around Jupiter.

Mass of Jupiter is given by

MJ = 42RIO3GTIO2 …. 1

Where,

MJ is the mass of Jupiter.

G is the Universal gravitational constant.

The orbital period of the Earth, Te = 365.25 days = 365.25 × 25 × 60 × 60s

The orbital radius of the Earth, Re = 1AU = 1.496 × 1011m

Mass of the Sun is given as

Ms=42Re3GTe2 …. 2

MSMJ=42Re3GTe2GTIO242RIO3=Re3RIO3TIO2Te2

MSMJ=1.769×24×60×60365.25×24×60×6021.496×10114.22×1083

MSMJ=1045.04

MSMJ~1000

We get,

MS ~ 1000 × MJ

The mass of Jupiter is therefore estimated to be approximately one thousandth that of the Sun.

Q2. A rocket is launched vertically at a speed of 5 km/s-1 from the surface of the Earth. How far does the rocket travel before coming back to Earth? Earth’s mass is 6.0 × 1024 kg; its mean radius is 6.4 × 106 m; and its gravitational constant is 6.67 × 10-11 Nm2kg-2.

Ans: Distance from the Earth’s centre = 8 × 106m

Velocity of the rocket, v = 5kms-1 = 5 × 103ms-1

Mass of the Earth, Me = 6.0 × 1024kg

Radius of the Earth, Re = 6.4 × 106m

Height reached by rocket mass m is h

At the surface of the Earth,

Total energy of the rocket = Kinetic energy + Potential energy

= 12mv2+-GMemRe

At highest point h, v = 0

Then, Potential energy = GMemRe+h

Total energy of the rocket = 0+GMemRe+h= –GMemRe+h

From the law of conservation,

Total energy of the rocket at the Earth’s surface = Total energy of rocket at height h

Then,

12mv2+-GMemRe=-GMemRe+h

12v2=GMe1Re1Re+h

12v2=GMeRe+h-ReRe(Re+h)

12v2=GMehReRe+hReRe

12v2=gRehRe+h

Where,

g = GMRe2=9.8ms-2 is the acceleration due to gravity on the Earth’s surface.

Clearly,

v2(Re+h) = 2gReh

v2Re = h (2gRe – v2)

h = Rev22gRev2

h = 6.4×25×1012100.44×106

h = 1.6 106m

Height achieved by the rocket with respect to the Earth’s centre, H is given by: H = Re + h

H = 6.4 106 + 1.6 106

We obtain,

H = 8.0 106m

The distance the rocket travels before landing on Earth is 8.0 106 metres.

Q3. A head-on collision is imminent between two stars, each of which has a solar mass of one solar mass (=2 1030 kg). Their speeds are barely perceptible at a distance of 109 kilometres. What is the rate of their collision? Each star has a 104 km radius. Assume that up until their collision, the stars won’t change. (Use the known value of G)

Ans: Mass of each star, M = 2 × 1030 kg

Radius of each star, R = 104 km = 107m

Distance between the stars, r = 109 km = 1012m

For negligible speeds, v = 0

Total energy of two stars separated at distance r is given by

TE=-GMMr+12mv2

TE= GMMr+0 ……(i)

Now, consider the case when the stars are about to collide:

Velocity of the stars = v

Distance between the centres of the stars = 2R

Total kinetic energy of both stars = 12Mv2+12Mv2=Mv2

Total potential energy of both stars = -GMM2R

Total energy of the two stars = Mv2GMM2R …..(ii)

From the law of conservation of energy,

Mv2GMM2R=-GMMR

v2=6.67×10-10×2×103011012+1107

v2=13.34×101910-12+5×10-8

v2=13.34×1019×5×10-8

v2=6.67×1012

v=6.67×1012

v=2.58×106m/s

The two stars collide at a speed of 2.58 × 106 m/s.

### Class 11 Physics Chapter 8 Important Questions

In order to familiarise yourself with the style, theme, depth, and techniques required to answer the questions in the exams, the set of Physics Class 11 Chapter 8 Important Questions from Extramarks is a helpful resource for exam preparation.

Importance of Class 11 Physics Chapter 8 Gravitation

Spending time studying Important Questions allows you to experiment with the best ways to respond to questions. You can improve your exam-taking skills by comparing your responses to the answers provided based on the official marking scheme. Practising these questions allows the following.

• Development of your time management skills
• Easier revision of main topics
• Focus on common exam themes
• Practise with actual exam-style questions

### Gravitation Class 11 Important Questions

Here are some of the important questions discussed of Chapter 8 Gravitation in Class 11 Physics.

Q1. the Earth is roughly spherical. What will be the acceleration caused by gravity on Earth’s surface if its interior includes materials that is not uniformly dense on all sides?

Options:

• Aimed at the Earth’s centre, yet it varies everywhere.
• Similar in value but not pointing at the centre
• Having a constant magnitude throughout and being pointed at the centre.
• Cannot at any time be zero.

Ans: Option (d) provides the correct response: “g” cannot be zero at any time.

Q2. Three particles are positioned at positions A, B, and C, respectively, with AB = (12) and masses 2M, m, and M. (BC)

At time t = 0, they are all at rest, and a body with mass “m” is much smaller than M. What will occur at later times before any collision occurs?

Options:

• Mass “m” is moving in the direction of mass “M.”
• The mass “m” is still at rest.
• “m” goes in the direction of “2M”.
• Movements oscillate from mass “m”.

Ans: Option (c): Mass “m” goes in the direction of mass “2M”.

### Class 11 Physics Chapter 8 Important Topics

• Escape Velocity
• Three laws of Kepler of planetary motion
• Universal Law of Gravitation
• Variation of the value of acceleration due to gravity ‘g’:
• With respect to depth
• With respect to the altitude
• Gravitational Potential and Gravitational Potential Energy
• Orbital Velocity
• Geo-stationary satellite
• Gravitation field intensity
• Mass and mean density of Earth
• Principle of superposition of gravitation

What You Will Get to Know on Extramarks?

Here’s why you should refer to the materials provided by Extramarks:

1. All study materials are prepared by subject matter experts.
2. The materials are regularly vetted and updated as per the latest CBSE Syllabus.
3. The solutions given are aligned with the revised CBSE assessment criteria.
4. You can easily access these materials from the Extramarks’ website at your convenience.

### Conclusion

Now that you understand the advantages of solving Important Questions Class 11 Physics Chapter 8, Extramarks recommends that you attempt these questions once every few days to remember the concepts and their applications for future questions. Also, keep in mind to practise these questions first before referring to the solutions given. To pass the test with flying colours, you must work both hard and intelligently.

Q.1 Read the assertion and reason carefully to mark the correct option out of the options given below.
Assertion: When a twin satellite straddles a region of increasing mass, with leading satellite leaving the region and the trailing satellite entering the region, the distance between the twin satellite changes.
Reason: The gravitational force acting between two bodies is proportional to the product of their mass and inversely proportional to the distance between them.

A-Assertion is true but reason is false.

B-Assertion and reason both are false.

C-Both assertion and reason are true and the reason is the correct explanation of the assertion.

D-Both assertion and reason are true but reason is not the correct explanation of the assertion.

Marks:1

Ans

Initially, when the leading satellite approaches a region of greater gravity, it is pulled a little bit farther ahead of the trailing satellite, which slightly increases the distance between both of them. After a certain time, as the lead satellite leaves the high-gravity area, it gets pulled slightly back while the trailing satellite, which is now approaching the high gravity region, is pulled slightly ahead, it decreases the gap between the two satellites.

Q.2 Gravitational force between two bodies is 1 Newton. If the distance between them is made twice, the force will now become

A-4 N.

B-3 N.

C-1/2 N.

D-1/4 N.

Marks:1

Ans

Q.3 The escape speed from the earth for a piece of 10 gram is 11.2 km/s. For a piece of 100 gram the speed will be

A-less than 11.2 km/s

B-more than 11.2 km/s

C-remains same

D-cannot be determined

Marks:1

Ans

Escape speed does not depend upon mass of the body to be escaped.

Q.4 What is the minimum energy required to launch a satellite of mass m from the surface of the earth of mass âMâ™ and radius â˜Râ™ in a circular orbit at an altitude 2R?

Marks:5

Ans

Energy of the satellite on the surface of the earth ,

Q.5 Find the depth below the surface of the earth where acceleration due to gravity becomes 30% of its value on the surface of the earth? Radius of the earth = 6.4 X 106m.

Marks:3

Ans

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### 1. Explain the term "gravitational potential".

The amount of effort required to move a unit mass without acceleration from infinity to a specific position is known as the gravitational potential.

### 2. Why does gravitational potential energy always have a negative value?

A mass is brought towards the centre of the Earth because of gravitational field. In other words, work must be done on a body if it is removed from the earth’s gravitational field. Therefore, the gravitational potential energy is negative on the surface of Earth. Since gravity is always attracting in nature, the potential energy of gravitation is always negative.

### 3. What are the most frequently asked questions from Class 11 Physics Chapter 8?

Class 11 Physics Chapter 8 covers one of the most crucial topics, ‘Gravity’. The following are the topics about which most questions are asked.

• Kepler’s Laws and the Universal Law of Gravitation
• Gravitational Potential Energy
• Characteristics of gravitational force
• Acceleration due to gravity

### 4. Name some of the important equations from Class 11 Physics Chapter 8.

Since Chapter 8 of Class 11 Physics concentrates on gravitational theory, the chapter’s key equations would follow closely behind all of these theories. Students can easily solve all questions after they comprehend the fundamentals and concepts underlying the formula. The following formulas are highlighted as being crucial:

• Acceleration due to gravity from a particular height h : ga=g1 – (2h/R)
• Acceleration due to gravity at a certain depth d : gd=g1 – (h/R)
• Gravitational field strength : E = F/m
• Relationship between G and g : g = Gm/R2
• Universal Law of Gravitation : F = Gm1m2/r2