NCERT Solutions for Class 11 Physics Chapter 8 Gravitation

NCERT Solutions Class 11 Physics Gravitation – Free PDF Download

NCERT Solutions for Class 11 Physics Chapter 8 Gravitation is a vital resource you must refer to to score good marks in the Class 11 term – I exam. This solution is similar to referring to a number of textbooks by the experts. These solutions strictly adhere to the latest update on the term – I CBSE Syllabus 2021-22. They present you the answers to the questions in the textbooks, important questions from previous year question papers and sample papers. The NCERT Solutions for Class 11 Physics Chapter 8 designed by the faculty at BYJU’S help students frame suitable answers for complex questions that appear in the first term exam.

In our lives, we know that all material objects are attracted towards the earth. Anything which we throw up falls down. Climbing a hill is tiring than going downhill. Want to know more about Gravitation? Not able to answer the questions from NCERT Textbook? Understanding the concepts are difficult? BYJU’S is the one-stop solution for all your needs. For more assistance, students can download the NCERT Solutions for Class 11 Physics Chapter 8 PDF from the link provided here.

Download NCERT Solutions Class 11 Physics Chapter 8 PDF:-Download Here

ncert solutions april9 class 11 physics chapter 8 gravitation 01
ncert solutions april9 class 11 physics chapter 8 gravitation 02
ncert solutions april9 class 11 physics chapter 8 gravitation 03
ncert solutions april9 class 11 physics chapter 8 gravitation 04
ncert solutions april9 class 11 physics chapter 8 gravitation 05
ncert solutions april9 class 11 physics chapter 8 gravitation 06
ncert solutions april9 class 11 physics chapter 8 gravitation 07
ncert solutions april9 class 11 physics chapter 8 gravitation 08
ncert solutions april9 class 11 physics chapter 8 gravitation 09
ncert solutions april9 class 11 physics chapter 8 gravitation 10
ncert solutions april9 class 11 physics chapter 8 gravitation 11
ncert solutions april9 class 11 physics chapter 8 gravitation 12
ncert solutions april9 class 11 physics chapter 8 gravitation 13
ncert solutions april9 class 11 physics chapter 8 gravitation 14
ncert solutions april9 class 11 physics chapter 8 gravitation 15

Access answers of NCERT Solutions for Class 11 Physics Chapter 8 Gravitation


Q.1: Answer the following :
(a) You can shield a charge from electrical forces by putting it inside a hollow conductor. Can you shield a body from the gravitational influence of nearby matter by putting it inside a hollow sphere or by some other means?
(b) An astronaut inside a small space ship orbiting around the earth cannot detect gravity. If the space station orbiting around the earth has a large size, can he hope to detect gravity?
(c) If you compare the gravitational force on the earth due to the sun to that due to the moon, you would find that the Sun’s pull is greater than the moon’s pull. (you can check this yourself using the data available in the succeeding exercises). However, the tidal effect of the moon’s pull is greater than the tidal effect of the sun. Why?
   

Solution:

(a). No, as of now, no method has been devised to shield a body from gravity because gravity is independent of medium and it is the virtue of each and every matter. So the shield would exert the gravitational forces.

(b). Yes, if the spaceship is large enough then the astronaut will definitely detect the Mars gravity.

(c). Gravitational force is inversely proportional to the square of the distance whereas, Tidal effects are inversely proportional to the cube of the distance. So as the distance between the earth and the moon is smaller than the distance between earth and the sun, the moon will have a greater influence on the earth’s tidal waves.

Q.2. Choose the correct alternative :
(a) Acceleration due to gravity increases/decreases with increasing altitude.
(b) Acceleration due to gravity increases/decreases with increasing depth (assume the earth to be a sphere of uniform density).
(c) Acceleration due to gravity is independent of the mass of the earth/mass of the body.
(d) The formula –G M m(1/r 2 – 1/r1) is more/less accurate than the formula mg(r2 – r1) for the difference of potential energy between two points r2 and r1 distance away from the centre of the earth.

Solution:

(a) decreases
(b) decreases
(c) mass of the body
(d) more

Q.3: Suppose there existed a planet that went around the Sun twice as fast as the earth. What would be its orbital size as compared to that of the earth?

Solution:

Time taken by the earth for one complete revolution, TE = 1 Year

Radius of Earth’s orbit, RE = 1 AU

Thus, the time taken by the planet to complete one complete revolution:

TP = 12\frac{1}{2}TE  = 12\frac{1}{2} year

Let, the orbital radius of this planet = RP

Now, according to the Kepler’s third law of planetary motion:

(RPRE)3=(TPTE)2\left ( \frac{R_{P}}{R_{E}} \right )^{3}=\left ( \frac{T_{P}}{T_{E}} \right )^{2} RPRE=(TPTE)23\\\frac{R_{P}}{R_{E}}=\left ( \frac{T_{P}}{T_{E}} \right )^{\frac{2}{3}}
RPRE=(TPTE)23=(12  TETE)23=(12)23=0.63\frac{R_{P}}{R_{E}}=\left ( \frac{T_{P}}{T_{E}} \right )^{\frac{2}{3}}=\left ( \frac{\frac{1}{2}\;T_{E}}{T_{E}} \right )^{\frac{2}{3}}=\left ( \frac{1}{2} \right )^{\frac{2}{3}}=0.63

Therefore, radius of orbit of this planet is 0.63 times smaller than the radius of orbit of the Earth.

Q.4: Io, one of the satellites of Jupiter, has an orbital period of 1.769 days and the radius of the orbit is 4.22 × 108 m. Show that the mass of Jupiter is about one-thousandth that of the sun.

Solution:

Given,

Orbital period of Io, TI0 = 1.769 days  =  1.769 × 24 × 60 × 60 s

Orbital radius of Io, RI0 = 4.22 × 108 m

We know the mass of Jupiter:

MJ = 4π2RI03 / GTI02    . . . . . . . . . . . . . . . (1)

Where;

MJ = Mass of Jupiter

G = Universal gravitational constant

Also,

The orbital period of the earth,

T= 365.25 days = 365.25 × 24 × 60 × 60 s

Orbital radius of the Earth, R= 1 AU = 1.496 × 1011 m

We know that the mass of sun is:

MS = 4  π2  RE3G  TE2\frac{4\;\pi ^{2}\;R_{E}^{3}}{G\;T_{E}^{2}}. . . . . . . . . . . . .  (2)

Therefore,  MSMJ\\\frac{M_{S}}{M_{J}} = 4  π2  RE3G  TE2  ×  G  T1024  π2  R103\frac{4\;\pi ^{2}\;R_{E}^{3}}{G\;T_{E}^{2}}\;\times \;\frac{G\;T_{10}^{2}}{4\;\pi ^{2}\;R_{10}^{3}}\\ = RE3TE2  ×  T102R103\frac{R_{E}^{3}}{T_{E}^{2}}\;\times \;\frac{T_{10}^{2}}{R_{10}^{3}}

Now, on substituting the values, we will get:

= [1.769×24×60×60365.25×24×60×60]2×[1.496×10114.22×108]3\left [ \frac{1.769\times 24\times 60\times 60}{365.25\times 24\times 60\times 60} \right ]^{2}\times \left [ \frac{1.496\times 10^{11}}{4.22\times 10^{8}} \right ]^{3} = 1045.04

Therefore, MSMJ\frac{M_{S}}{M_{J}} ~ 1000

M~ 1000 × MJ     

[Which proves that, the Sun’s mass is 1000 times that of Jupiter’s]

Q.5. Let us assume that our galaxy consists of 2.5 × 1011 stars each of one solar mass. How long will a star at a distance of 50,000 ly from the galactic centre take to complete one revolution? Take the diameter of the Milky Way to be 105ly. 

Solution:

Mass of our galaxy, M=2.5×10 11 solar mass
1 Solar mass = Mass of Sun =2×1030 kg
Mass of our galaxy, M=2.5×1011×2×10 30=5×1041kg

Diameter of Milky Way,d=105 ly
Radius of Milky Way,r=5×10ly
1 ly=9.46×1015m

Therefore, r=5×104×9.46×1015=4.73×1020m

The time taken by the star to revolve around the galactic centre is given by the relation:
T=(4π2r3/GM)1/2
= 4×(3.14)2×(4.73)3×10606.67×1011×5×1041\sqrt{\frac{4\times (3.14)^{2}\times (4.73)^{3}\times 10^{60}}{6.67\times 10^{-11}\times 5\times 10^{41}}}

= 1.12 x 1016 s

1.12×1016365×24×60×60years\frac{1.12\times 10^{16}}{365\times 24\times 60\times 60}\, years

= 3.55 x 108 years

Q.6. Choose the correct alternative:
(a) If the zero of potential energy is at infinity, the total energy of an orbiting satellite is negative of its kinetic/potential energy.
(b) The energy required to launch an orbiting satellite out of earth’s gravitational influence is more/less than the energy required to project a stationary object at the same height (as the satellite) out of earth’s influence.

Solution:

(a) If the zero potential energy is at infinity, the total energy of an orbiting satellite is negative of its kinetic energy.
(b) The energy required to launch an orbiting satellite out of Earth’s gravitational influence is less than the energy required to project a stationary object at the same height (as the satellite) out of Earth’s influence.

Q. 7. Does the escape speed of a body from the earth depend on (a) the mass of the body, (b) the location from where it is projected, (c) the direction of projection, (d) the height of
the location from where the body is launched?

Solution:

The escape speed is given by the expression

v=2GMR=2gRv= \sqrt{\frac{2GM}{R}}=\sqrt{2gR}

(a) The escape speed of a body from the Earth does not depend on the mass of the body.
(b) The escape speed of a body from the Earth does not depend on the location from where a body is projected.
(c) The escape speed does not depend on the direction of projection of a body.
(d) The escape speed of a body depends upon the height of the location from where the body is launched since the escape velocity depends on the gravitational potential at the point from which it is launched. This potential in turn depends on the height.

Q.8. A comet orbits the Sun in a highly elliptical orbit. Does the comet have a constant (a) linear speed (b) angular speed (c) angular momentum (d) kinetic energy (e) potential energy (f) total energy throughout its orbit? Neglect any mass loss of the comet when it comes very close to the Sun.
Solution:

(a) According to Kepler’s second law the linear speed of the comet keeps changing. When the comet is near the sun, its speed will be the fastest and when it is far away from the sun, its speed will be the least.
(b) Angular speed also varies slightly.
(c) Comet has constant angular momentum.
(d) Kinetic energy changes
(e) Potential energy changes along the path.
(f) Total energy will remain constant throughout the orbit

 

Q9. Which of the following symptoms is likely to afflict an astronaut in space (a) swollen feet, (b) swollen face, (c) headache, (d) orientational problem

Solution:

(a). In zero gravity the blood flow to the feet isn’t increased so the astronaut does not get swollen feet.

(b). There is more supply of blood to the face of the astronaut. Therefore, the astronaut will have a swollen face

(c). Due to increased blood supply to their faces, astronauts can be affected by headaches.

(d). Space has different orientations, so orientational problems can affect an astronaut.

Q.10: In the following two exercises, choose the correct answer from among the given ones:
The gravitational intensity at the centre of a hemispherical shell of uniform mass density has the direction indicated by the arrow (see Fig ) (i) a, (ii) b, (iii) c, (iv) 0

 

NCERT Solutions for Class 11 Physics Chapter 8 Gravitation Question 10

Solution: (iii) c

Reason:

Inside a hollow sphere, gravitational forces on any particle at any point is symmetrically placed. However, in this case, the upper half of the sphere is removed. Since gravitational intensity is gravitational force per unit mass it will act in a direction point downwards along ‘c’.

Q.11: For the above problem, the direction of the gravitational intensity at an arbitrary
point P is indicated by the arrow (i) d, (ii) e, (iii) f, (iv) g.

Solution: (ii) e

Reason:

Making use of the logic/explanation from the above answer we can conclude that the gravitational intensity at P is directed downwards along e

Q.12. A rocket is fired from the earth towards the sun. At what distance from the earth’s centre is the gravitational force on the rocket zero? Mass of the sun = 2 x 1030 kg, mass of the earth = 6 x 1024 kg. Neglect the effect of other planets etc. (orbital radius = 1.5 x 1011 m).

Solution:

Mass of Sun, Msun= 2 x 1030 kg

Mass of Earth, Mearth = 6 x 1024 kg

Orbital radius, r = 1.5 x 1011 m

Mass of the rocket = m

Let P be the point at a distance x at which the  gravitational force on the rocket due to Earth

Applying Newton’s law of gravitation, the gravitational force acting on the rocket at point P is equated.
GmMsun(rx)2=GmMearthx2\frac{GmM_{sun}}{(r-x)^{2}}=\frac{GmM_{earth}}{x^{2}}
(rx)2x2=MsunMearth=(2×103060×1024)1/2=577.35\frac{(r-x)^{2}}{x^{2}}=\frac{M_{sun}}{M_{earth}}= \left ( \frac{2\times 10^{30}}{60\times 10^{24}} \right )^{1/2}=577.35
1.5 x 1011 – x = 577.35x
x = 1.5 x 1011/578.35 = 2.59 x 108 m

Q.13: How will you ‘weigh the sun’, that is estimate its mass? The mean orbital radius of the earth around the sun is 1.5 × 108 km

Solution:

Given:
Earth’s orbit, r = 1.5 × 1011 m
Time taken by the Earth for one complete revolution,
T = 1 year = 365.25 days
i.e. T = (365.25 × 24 × 60 × 60) seconds

Since, Universal gravitational constant, G = 6.67 × 10–11 Nm2 kg–2
Therefore, mass of the Sun, M = 4  π2r3G  T2\frac{4\;\pi ^{2}r^{3}}{G\;T^{2}}
M=4×(3.14)2×(1.5×1011)3(6.67×1011)×(365.25×24×60×60)2\\\Rightarrow M =\frac{4\times (3.14)^{2}\times (1.5\times 10^{11})^{3}}{(6.67\times 10^{-11})\times (365.25\times 24\times 60\times 60)^{2}} M=4×(3.14)2×(1.5×1011)3(6.67×1011)×(365.25×24×60×60)2\\\Rightarrow M =\frac{4\times (3.14)^{2}\times (1.5\times 10^{11})^{3}}{(6.67\times 10^{-11})\times (365.25\times 24\times 60\times 60)^{2}} \\\Rightarrow   M=1.331×103566.425×103=2.004×1030  kgM=\frac{1.331\times 10^{35}}{66.425\times 10^{3}}=2.004\times 10^{30}\;kg
Therefore, the estimated mass of the Sun is 2.004 × 1030 Kg

Q. 14. A Saturn year is 29.5 times the Earth year. How far is the Saturn from the Sun if the Earth is 1.50 x 108 km away from the Sun?

Solution:

According to Kepler’s third law of planetary motion

T=4π2r3GMT =\sqrt{\frac{4\pi ^{2}r^{3}}{GM}}

From the above equation, we get T2 ∝r3

Ts and rs is the orbital period and the mean distance of Saturn from the Sun respectively.

Te and re is the orbital period and the mean distance of Earth from the Sun respectively.

Time period of Saturn ,Ts = 29.5 Te

⇒rs3/re3 = Ts2/Te2

rs=re(Ts/Te)2/3

= 1.5 x 10x 29.52/3

= 14.3 x 10km

Q.15: A body weighs 63 N on the surface of the earth. What is the gravitational force on it due to the earth at a height equal to half the radius of the earth?

Solution:

Given:

Weight of the man, W = 63 N

We know that acceleration due to gravity at height ‘h’ from the Earth’s surface is:

g’ = g[1+(hRe)]2\frac{g}{\left [ 1+\left ( \frac{h}{R_{e}} \right ) \right ]^{2}}
Where, g = Acceleration due to gravity on the Earth’s surface
And, Re = Radius of the Earth
For h = Re2\frac{R_{e}}{2}
g’ = g[1+(Re2Re)]2\frac{g}{\left [ 1+\left ( \frac{R_{e}}{2R_{e}} \right ) \right ]^{2}}
\\\Rightarrow g’ = g[1+(12)]2=49  g\frac{g}{\left [ 1+\left (\frac{1}{2} \right ) \right ]^{2}}=\frac{4}{9}\;g

Also , the weight of a body of mass ‘m’ kg at a height of ‘h’ meters can be represented as :
W’ = mg
= m × 49\frac{4}{9}g  =  49\frac{4}{9}mg
= 49\frac{4}{9}W
= 49\frac{4}{9} × 63  = 28 N.

Q. 16. Assuming the earth to be a sphere of uniform mass density, how much would a body weigh halfway down to the centre of the earth if it weighed 250 N on the surface?

Solution:

Weight of a body on the Earth’s surface, W=mg=250 N

Radius of the Earth = Re
Let d be at a distance halfway to the centre of the earth, d=Re/2

Acceleration due to gravity at d is given by the relation
gd=(1− d/Re)g
gd=(1−Re/2Re)g​
=g/2

Weight of the body at depth d,
W′=mgd
=mg/2 = W/2 = 250/2
=125 N

Q.17: A rocket is fired vertically with a speed of 5 km s-1 from the earth’s surface. How far from the earth does the rocket go before returning to the earth? Mass of the earth = 6.0 × 1024 kg; mean radius of the earth = 6.4 × 106 m; G = 6.67 × 10–11 N m2 kg–2

Solution:

Given:
Velocity of the missile,  v = 5 km/s = 5 × 103 m/s
Mass of the Earth, ME = 6 × 1024 kg
Radius of the Earth, RE = 6.4 × 106 m

Let, the height reached by the missile be ‘h’ and the mass of the missile be ‘m’.
Now, at the surface of the Earth:
The total energy of the rocket at the surface of the Earth = Kinetic energy + Potential energy
TE1 = 12mv2+GME  mRE\frac{1}{2}mv^{2}+\frac{-GM_{E}\;m}{R_{E}}

Now, at highest point ‘h’:

Kinetic Energy = 0   [Since, v = 0]
And, Potential energy = GME  mRE+h\frac{-GM_{E}\;m}{R_{E}+h}

Therefore, total energy of the missile at highest point ‘h’:

TE2 = 0 + GME  mRE+h\frac{-GM_{E}\;m}{R_{E}+h} \Rightarrow TE2 = GME  mRE+h\frac{-GM_{E}\;m}{R_{E}+h}

According to the law of conservation of energy, we have :
Total energy of the rocket at the Earth’s surface TE1 = Total energy at height ‘h’ TE2:
12mv2+GME  mRE=GME  mRE+h12v2+GMERE=GMERE+hv2=2GME×[1RE1(RE+h)]=2GME[h(RE+h)RE]\Rightarrow \frac{1}{2}mv^{2}+\frac{-GM_{E}\;m}{R_{E}}=\frac{-GM_{E}\;m}{R_{E}+h}\\ \Rightarrow \frac{1}{2}v^{2}+\frac{-GM_{E}}{R_{E}}=\frac{-GM_{E}}{R_{E}+h}\\ \Rightarrow v^{2} = 2GM_{E}\times \left [ \frac{1}{R_{E}}-\frac{1}{(R_{E}+h)} \right ]=2GM_{E}\left [ \frac{h}{(R_{E}+h)R_{E}} \right ]

v2 = GMEh/RE2 +REh

Where, g=GMRE2\\g=\frac{GM}{R_{E}^{2}} = 9.8 ms-2

V2 = 2gRE2h/RE2+REh
v2=2g  RE  hRE+h\Rightarrow v^{2}=\frac{2g\;R_{E}\;h}{R_{E}+h}

Therefore, v2 (RE + h) = 2gREh
\\\Rightarrow v2RE = h (2gRE – v2)

\\\Rightarrow h=RE  v22gREv2=(6.4×106)×(5×103)22×9.8×6.4×106(5×103)2h=\frac{R_{E}\;v^{2}}{2gR_{E}-v^{2}}=\frac{(6.4\times 10^{6})\times (5\times 10^{3})^{2}}{2\times 9.8\times 6.4\times 10^{6}-(5\times 10^{3})^{2}}
\\\Rightarrow h = 1.6 × 10 m
Therefore, the missile reaches at height of 1.6 × 10m from the surface.

Now, the distance from the center of the earth = h + RE = 1.6 × 10+ 6.4 x 106 = 8.0  x106 m

Therefore, the greatest height the missile can attain before falling back to the earth is 1.6 × 106   m and the final distance of the missile from the center of the earth is 8.0  x106 m

Q.18. The escape speed of a projectile on the earth’s surface is 11.2 km s–1. A body is projected out with thrice this speed. What is the speed of the body far away from the earth? Ignore the presence of the sun and other planets.

Solution:

Given,
Escape speed of the projectile on the Earth’s surface, Ve = 11.2 km/s
Speed of projection of the body, v = 3Ve = 3×11.2 = 33.6 km/s
Let v and v’ be the speed of the body at the time of projection and at a point far from the earth.
At the time of projection
Initial kinetic energy of the body = 1/2mv²
Initial gravitational potential energy of the body = -GMem/Re
Me is the mass of the Earth

Re is the radius of the Earth

At a larger distance from the earth surface 
KE of the body = 1/2mv’²
The gravitational potential energy of the body = 0

According to law of conservation of energy
Total energy at the point of projection = total energy at very far from the earth’s surface.
1/2mv² + (-GMem/Re) = 1/2mv’²
1/2 mv’² = 1/2mv² – GMem/Re——–(1)
If Ve is the escape velocity , then,
1/2 mVe² = GMem/R———(2)

Substituting  (2) in (1)
1/2mv’² = 1/2mv² – 1/2mVe²
v’² = v – Ve²
= (3Ve)² – Ve²
= 8Ve²
v’² = 8Ve²
v’ = √8 x Ve
v = 2 × 1.414 × 11.2 km/s
= 31.68 km/s
Speed of the body far away from the earth = 31.68 km/s
Q. 19. A satellite orbits the earth at a height of 400 km above the surface. How much energy must be expended to rocket the satellite out of the earth’s gravitational influence? Mass of the satellite = 200 kg; mass of the earth = 6.0×1024 kg; radius of
the earth = 6.4 × 10m; G = 6.67 × 10–11 N m2 kg–2
Solution:
Height of the satellite, h = 400 km = 4 × 105 m
Mass of the Earth, M = 6.0 × 10²⁴ kg
Mass of the satellite, m = 200 kg
Radius of the Earth, Re = 6.4 × 106 m
Total energy of the satellite at height h = kinetic energy + potential energy
= (1/2)mv²  + [ -GmM/ (Re + h) ]
Orbital velocity of the satellite, v=GMeRe+hv = \sqrt{\frac{GM_{e}}{R_{e}+h}}
Total energy of height, h = =12GMemRe+hGMemRe+h= \frac{1}{2}\frac{GM_{e}m}{R_{e}+h}-\frac{GM_{e}m}{R_{e}+h}

Total Energy= 12GMemRe+h- \frac{1}{2}\frac{GM_{e}m}{R_{e}+h}

The negative sign indicates that the satellite is bound to the earth. This is known as the binding energy of a satellite.

The energy required to send the satellite out of its orbit = – (Bound energy)

12GMemRe+h\frac{1}{2}\frac{GM_{e}m}{R_{e}+h}

Putting values of all terms.

=6.67×1011×6×1024×2002(6.4×106+4×105)\frac{6.67 \times 10^{-11}\times 6\times 10^{24}\times 200}{2(6.4\times 10^{6}+4\times 10^{5})}

= 5.9 × 109 J
Q. 20. Two stars each of one solar mass (= 2×1030 kg) are approaching each other for a head-on collision. When they are a distance 10km, their speeds are negligible. What is the speed with which they collide? The radius of each star is 104
km. Assume the stars to remain undistorted until they collide. (Use the known value of G).
Solution:
Mass of each star, M = 2 × 1030 kg
Radius of each star, R = 104 km = 107 m
Distance between the stars, r = 109 km = 1012m
Initial potential energy of the system = -GMm/r ——–(1)
Total K.E. of the stars = 1/2Mv2 + 1/2Mv= Mv2
where v is the speed of stars with which they collide. The distance between the centres of the stars at collision r = 2 R.
Therefore, the final potential energy of two stars when they are close to each other = -GMm/2R
Thus, the total energy of the two stars just before collision
E = Mv-GMm/2R
Since the energy is conserved, the initial energy is equal to the final energy
Mv-GMm/2R = -GMm/r
v2 =(Gm/2R) – (Gm/r)
= Gm (1/2R – 1/r)
= 6.67 x 10-11 x 2 x 1030 [(1/2 x 107) – (1/1012)]
v2= 6.67 x 1012
v =  2.58 x 106 m/s
Q.21: Two heavy spheres each of mass 100 kg and radius 0.10 m are placed 1.0 m apart on a horizontal table. What is the gravitational force and potential at the midpoint of the line joining the centres of the spheres? Is an object placed at that point in equilibrium? If so, is the equilibrium stable or unstable?

Solution:

Given:

Radius of spheres, R = 0.10 m

Distance between two spheres, r = 1.0 m

Mass of each sphere, M = 100 kg

NCERT Chapter 8 Class 11 Question 21

From the above figure, ‘A’ is the mid-point and since each sphere will exert the gravitational force in the opposite direction. Therefore, the gravitational force at this point will be zero.

Gravitational potential at the midpoint (A) is;

U= [GMr2+GMr2]\left [ \frac{-GM}{\frac{r}{2}}+\frac{-GM}{\frac{r}{2}} \right ]

U= [4GMr]\left [ \frac{-4GM}{r} \right ]

U= [4×(6.67×1011)×(1000)1.0]\left [ \frac{-4\times (6.67\times 10^{-11})\times (1000)}{1.0} \right ] \Rightarrow U= -2.668 x 10-7 J /kg

Therefore, the gravitational potential and force at the mid-point of the line connecting the centres of the two spheres is = -2.668 x 10-7 J /kg

The net force on an object, placed at the mid-point is zero. However, if the object is displaced even a little towards any of the two bodies it will not return to its equilibrium position. Thus, the body is in an unstable equilibrium.

Q.22: As you have learnt in the text, a geostationary satellite orbits the earth at a height of nearly 36,000 km from the surface of the earth. What is the potential due to earth’s gravity at the site of this satellite? (Take the potential energy at infinity to be zero). Mass of the earth = 6.0×1024 kg, radius = 6400 km

Solution:

Given:

Radius of the Earth, R = 6400 km = 0.64 × 107 m

Mass of Earth, M = 6 x 1024 kg

Height of the geostationary satellite from earth’s surface, h = 36000 km = 3.6 x 107 m

Therefore, gravitational potential at height ‘h’ on the geostationary satellite due to the earth’s gravity:

GP = GMR+h\frac{-GM}{R+h}
\\\Rightarrow GP = (6.67×1011)×(6×1024)(0.64×107)+(3.6×107)\frac{-(6.67\times 10^{-11})\times (6\times 10^{24})}{(0.64\times 10^{7})+(3.6\times 10^{7})}
\\\Rightarrow GP = 40.02×10134.24×107\frac{-40.02\times 10^{13}}{4.24\times 10^{7}}= -9.439 × 106 J/Kg

Therefore, the gravitational potential due to Earth’s gravity on a geostationary satellite orbiting earth is -9.439 × 106 J/Kg

Q.23: A star 2.5 times the mass of the sun and collapsed to a size of 12 km rotates with a speed of 1.2 rev. per second. (Extremely compact stars of this kind are known as neutron stars. Certain stellar objects called pulsars belong to this category). Will an object placed on its equator remain stuck to its surface due to gravity? (mass of the sun = 2×1030 kg).   

Solution:

Any matter/ object will remain stuck to the surface if the outward centrifugal force is lesser than the inward gravitational pull.

Gravitational force, fG = GM  mR2\frac{GM\;m}{R^{2}} [Neglecting negative sign]

Here,
M = Mass of the star = 2.5× 2 × 1030 = 5 × 1031 kg
m = Mass of the object
R = Radius of the star = 12 km = 1.2 ×104 m
Therefore, fG = (6.67×1011)×(5×1031)  m(1.2×104)2=1.334×1013m  N\frac{(6.67\times 10^{-11})\times (5\times 10^{31})\;m}{(1.2\times 10^{4})^{2}}=1.334\times 10^{13}m\;N

Now, Centrifugal force, fc= m r ω2
Here, ω = Angular speed = 2πν
ν = Angular frequency = 10 rev s–1
fc = m R (2πν)2

fc = m × (1.2 x 104) × 4 × (3.14)2 × (1.2)2 = (6.82 ×105m) N

As fG > fC, the object will remain stuck to the surface of black hole.

Q. 24.  A spaceship is stationed on Mars. How much energy must be expended on the spaceship to launch it out of the solar system? Mass of the space ship = 1000 kg; mass of the sun = 2×1030 kg; mass of mars = 6.4×1023 kg; radius of mars = 3395 km; radius of the orbit of mars = 2.28 ×10km; G = 6.67×10-11 N m2kg–2.

Solution:

Mass of the spaceship, ms =1000 kg
Mass of the Sun, M=2×1030kg
Mass of Mars, mm =6.4×1023kg
Radius of orbit of Mars, R=2.28×1011 m
Radius of Mars, r=3395×103m
Universal gravitational constant, G=6.67×10−11Nm2kg−2

The potential energy of the spaceship due to the gravitational attraction of the Sun, Us =−GMms/R
Potential energy of the spaceship due to the gravitational attraction of Mars, Um=−Gmmms/r

Total energy of the spaceship, E= Um+Us =[−GMms/R]+[-Gmmms/r]

The negative sign indicates that the satellite is bound to the system.
Energy required to launch the spaceship out of the solar system =−(total energy of the spaceship)

– E = [GMms/R]+[Gmmms/r]

= Gms(M/R + mm/r)

=  6.67×1011×103×(2×10302.28×1011+6.4×10233.395×106)6.67 \times 10^{-11}\times 10^{3}\times \left ( \frac{2\times 10^{30}}{2.28\times 10^{11}}+\frac{6.4 \times 10^{23}}{3.395\times 10^{6}} \right )

= 5. 91 x 1011 J

Q.25. A rocket is fired ‘vertically’ from the surface of mars with a speed of 2 km s–1. If 20% of its initial energy is lost due to martian atmospheric resistance, how far will the rocket go from the surface of mars before returning to it? Mass of mars = 6.4×1023 kg; radius of mars = 3395 km; G = 6.67×10-11 N m2 kg–2.

Solution:

Speed of the rocket fired from the surface of mars ( v) = 2 km/s

Mass of the rocket = m
Mass of the Mars ( M) = 6.4 × 10²³ Kg
Radius of the Mars (R) = 3395 km = 3.395 × 106 m

Initial Potential energy = – GMm/R
Initial kinetic energy of the rocket = 1/2 mv²

Total initial energy = 1/2 mv² – GMm/R

Due to atmospheric resistance, 20% of the kinetic energy is lost by the rocket.
The remaining Kinetic energy= 80% of 1/2mv²
= 2/5 mv²= 0.4 mv² ——–(1)

When the rocket reaches the highest point, at the height h above the surface the kinetic energy will be zero and the potential energy is equal to – GMm/(R+h)

Applying Law of conservation of energy.
– GMm/(R+h) =  – GMm/R + (0.4) mv²

GM/(R+h) =  1/R [ GM – 0.4Rv2]
(R+h/R) = GM/[ GM – 0.4Rv2]

h/R = {GM/[ GM – 0.4Rv2]} – 1

h/R = (0.4Rv2/GM – 0.4Rv2)

h = (0.4R2v2/GM – 0.4Rv2)

h=0.4×(2×103)×(3.395×106)26.67×1011×6.4×10230.4×(2×103)2×3.395×106h = \frac{0.4\times (2\times 10^{3})\times (3.395\times 10^{6})^{2}}{6.67\times 10^{-11}\times 6.4\times 10^{23}-0.4\times (2\times 10^{3})^{2}\times 3.395\times 10^{6}}

h = 495 km

NCERT Solutions help your preparation with worksheets, exemplar problems, MCQs (multiple choice questions), HOTS (high order thinking skills), short answer questions, NCERT tips and tricks. Solving this solution repeatedly will help you in understanding the types of questions asked in Class 11 term-wise examinations and entrance exams for the under-graduation course.

Class 11 Physics NCERT Solutions for  Chapter 8 Gravitation

Gravitation is a very popular subject for Class 11 students as most of the topics you study in the future are based on this phenomenon. We should be knowing the difference between gravitation and gravity in order to understand more complex subjects. This chapter consists of questions belonging to various important concepts such as how can a body be shielded from gravitational influences of any nearby matter, whose gravitational force is greater on earth, the sun’s or the moon’s.

We find various complex but easy to understand topics in this chapter such as acceleration due to gravity, finding potential energy difference between two points which are at a certain distance from the earth’s centre. We can even know that the acceleration due to gravity increases/decreases when the depth increases or altitude decreases. We will be finding questions on an interplanetary motion such as a planet revolving around the sun with a speed double than that of earth, this question will show you the use of Kepler’s laws of motion. We will be seeing questions on Jupiter’s satellite radius of revolutions.

This chapter talks about the distance and speed of light among stars which are far away in a different galaxy and the time taken for the stars to complete one revolution. Here, we shall see questions related to a space shuttle escape velocity and its dependency on factors like an object’s mass, location, direction of projection and the gravitational influences on it. We will be finding the angular speed, kinetic energy, total energy, potential energy, linear speed and angular momentum of an asteroid revolving around a star.

We will be going through the medical problems faced by astronauts in space and the reason behind it. We will be seeing problems with gravitational intensity within a hemisphere. We will be knowing the mass of the sun in one of the questions asked below. Do you know the distance between Uranus and earth and the time taken to reach there? You will get to know it by studying this solution.

We will see here the maximum height attained by a missile when it is fired vertically upwards before falling down on the earth’s surface. We will see the international space station’s motion around the earth and the influence of the earth’s gravitational field upon it. Do you know what will happen when two planets collide with each other, what will be their speed? We will see it here. How a star turns into a black hole and the energy required to launch a space station in mars are the type of questions you will find below.

In order to help you to cover this topic in a detailed and comprehensive way by solving problems, we have provided NCERT Solutions for Class 11 Physics Chapter 8 PDF to help students to learn better.

Important Concepts of NCERT Solutions for Class 11 Physics Chapter 8

Chapter 8 provides you all the required knowledge about Gravitation and related important concepts. The NCERT Solutions provides clear understanding of the fundamental concepts. Some of the important concepts explained in the Chapter 8 are –

8.1 Introduction
8.2 Kepler’s laws
8.3 Universal law of gravitation
8.4 The gravitational constant
8.5 Acceleration due to gravity of the earth
8.6 Acceleration due to gravity below and above the surface of earth
8.7 Gravitational potential energy
8.8 Escape speed
8.9 Earth satellites
8.10 Energy of an orbiting satellite
8.11 Geostationary and polar satellites
8.12 Weightlessness

Frequently Asked Questions on NCERT Solutions for Class 11 Physics Chapter 8

Does BYJU’S website provide free NCERT Physics Solutions for Class 11 Chapter 8 Gravitation?

Yes, these solutions can be viewed and downloaded in free PDF format. Students who find it difficult in solving exercise wise questions of NCERT Physics for Class 11 Chapter 8 Gravitation can access PDF solutions. The questions are solved by our experts in the best possible way to make it easier for the students during the first term exam preparation.

Explain gravitational force covered in Chapter 8 of NCERT Solutions for Class 11 Physics.

According to Newton’s universal law of gravitation, the force of attraction between any two bodies is directly proportional to the product of their masses and is inversely proportional to the square of the distance between them. Students who wish to understand the concepts covered in this chapter are advised to download the PDF of solutions from BYJU’S. The solutions are created by a set of highly experienced subject matter experts with the aim of helping students to score well in the term – I exams.

Why are the NCERT Solutions for Class 11 Physics Chapter 8 important for the term – I exams?

Class 11 is considered to be one of the crucial steps in students’ lives. For this purpose, the faculty at BYJU’S has designed the chapter-wise solutions after conducting vast research on each concept. The solutions contain detailed and stepwise explanations to help students score well in the first term exams. PDF format of solutions is available on BYJU’S which can be downloaded and used by the students based on their requirements.

Leave a Comment

Your Mobile number and Email id will not be published. Required fields are marked *

*

*