NCERT Solutions for Class 11 Physics Chapter 8 Gravitation

NCERT solutions class 11 Physics Chapter 8 Gravitation is a vital resource you must refer to score good marks in class 11 examination. This solution is the result of referring to a number of textbooks by experts. These solutions are prepared as per the latest CBSE syllabus 2018-19. They present you the answers to the question in the textbooks, important questions from previous year question papers and sample papers.

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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 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 if greater on earth, the suns or moons?.

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 satellites 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. We will see here questions related to a space shuttles escape velocity and its dependency on factors like an object’s mass, location and 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 on 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.

Important questions of 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 sun. Why?
   

Sol:

(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 moon is smaller than the distance between earth and sun, the moon will have a greater influence on the earth’s tidal waves.

Q2. 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 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.

Sol:

(a). body.

(b). more.

(c). decreases.

(d). increases.

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?

Sol:

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.

 Sol:

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 mass of Jupiter:

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

Where;

MJ = Mass of Jupiter

G = Universal gravitational constant

Also,

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 105 ly

 Sol:

Mass of Sombrero Galaxy, M = 3 × 1011 solar mass
Solar mass = Mass of Sun = 2.0 × 1036 kg
Mass of the galaxy, M = 3 × 1011 × 2 × 1036 = 6 × 1041 kg
Diameter of Sombrero Galaxy, d = 5 x 104 ly
Radius of Sombrero Galaxy, r = 2.5 × 104 ly

We know that:

1 light year = 9.46 × 1015 m
Therefore, r = 2.5 × 104 × 9.46 × 1015= 2.365 x 1020 m

As this star revolves around the massive black hole in center of the Sombrero galaxy, its time period can be found with the relation:
T = [4  π2  r3GM]12\left [ \frac{4\;\pi ^{2}\;r^{3}}{GM} \right ]^{\frac{1}{2}}\\
[4×3.142×(2.365×1020)3(6.67×1011)×(6×1041)]12\\\left [ \frac{4\times 3.14^{2}\times (2.365\times 10^{20})^{3}}{(6.67\times 10^{-11})\times (6\times 10^{41})} \right ]^{\frac{1}{2}}\\ =   4.246 x 1015  s
Now we know, 1 year = 365 × 24 × 60 × 60 s

1s = 1365×24×60×60\frac{1}{365\times 24\times 60\times 60}\\ years
Therefore, 4.246 × 1015s = 4.246×1015365×24×60×60\frac{4.246\times 10^{15}}{365\times 24\times 60\times 60} =  1.34 × 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

Sol:

(a).  more

(b).  kinetic.

Q7. 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?

Sol: (b)

Explanation:

Escape velocity is independent of the direction of projection and the mass of the body. It depends upon the gravitational potential at the place from where the body is projected. Gravitational potential depends slightly on the altitude and the latitude of the place, thus escape velocity depends slightly upon the location from where it is projected.

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.

Sol:

An asteroid orbiting a star will have constant angular momentum and the constant value of total energy throughout its 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

Sol:

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

(b). Due to zero gravity, the weight the bones have to bear is greatly reduced, this causes bone loss in astronauts spending greater amounts of time in space.

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

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

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 8.12) (i) a, (ii) b, (iii) c, (iv) 0

 

hemispherical sphere

Sol: (i) 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 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.

Sol: (iii) m

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

Q12. 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 = 2×1030 kg, mass of the earth = 6×1024 kg. Neglect the effect of other planets etc. (orbital radius = 1.5 × 1011 m).

Sol:

Given:

Mass of the comet, MC = 7.4 × 1024 kg

Mass of the Earth, ME = 6 × 10 24 kg
Orbital radius, r = 3.84 × 1010 m
Mass of the shuttle = m kg

Let ‘x’ be the distance from the center of the Earth where the gravitational force acting on the Shuttle ‘S’ becomes zero.

According to Newton’s law of gravitation, we have:

G  m  MA(rx)2=G  m  MEx2\\\Rightarrow \frac{G\;m\;M_{A}}{(r-x)^{2}}=\frac{G\;m\;M_{E}}{x^{2}}\\ \\\Rightarrow    MAME=(rxx)2\frac{M_{A}}{M_{E}}=\left ( \frac{r-x}{x} \right )^{2} \Rightarrow   rx1=(7.4×10246×1024)12\frac{r}{x}-1=\left ( \frac{7.4\times 10^{24}}{6\times 10^{24}} \right )^{\frac{1}{2}}\\ \Rightarrow   rx1=1.1106\frac{r}{x}-1=1.1106 \\\Rightarrow   x=r2.1106=3.84×10102.1106=1.819×1010  mx=\frac{r}{2.1106}=\frac{3.84\times 10^{10}}{2.1106}=1.819\times 10^{10}\;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

Sol:

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

Q14. A saturn year is 29.5 times the earth year. How far is the saturn from the sun if the earth is 1.50 × 108 km away from the sun ?

Sol:

Given:

Distance between Earth and the Sun, re = 15 × 107 km = 1.50 × 108 m
Time period of the Earth = T­e
Time period of Uranus, Tu = 84 Te

Let, the distance between the Sun and the Saturn be r

Now, according to the Kepler’s third law of planetary motion:
T =(4π2r3GM)12\left ( \frac{4\pi ^{2}r^{3}}{GM} \right )^{\frac{1}{2}}

For Uranus and Sun, we can write:
ru3re3=Tu2Te2ru=re[TuTe]23ru=1.5×1011[84  TeTe]23=1.5×1011×(84)23\Rightarrow \frac{r_{u}^{3}}{r_{e}^{3}}=\frac{T_{u}^{2}}{T_{e}^{2}}\\\\\\ \Rightarrow r_{u}=r_{e}\left [ \frac{T_{u}}{T_{e}} \right ]^{\frac{2}{3}}\\\\\\ \Rightarrow r_{u}=1.5\times 10^{11}\left [ \frac{84\;T_{e}}{T_{e}} \right ]^{\frac{2}{3}}=\boldsymbol{1.5\times 10^{11}\times (84)^{\frac{2}{3}}} ru=28.77×1012\Rightarrow r_{u}=28.77\times 10^{12}\\m
Therefore, Saturn is  28.77 × 1012 m away from the Sun.

 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?

Sol:

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?

Sol:

Given,

The weight of a body having mass ‘m’ at the surface of earth, W = mg = 250 N
Body of mass m’ is located at depthd = 13\frac{1}{3} Re
Where, Re = Radius of the Earth 
Now, acceleration due to gravity g’ at depth (d) is given by the relation:
g’ = 1(dRe)g1-\left ( \frac{d}{R_{e}} \right )g\\
g=1(Re3Re)g=23  g\Rightarrow g’ = 1-\left ( \frac{R_{e}}{3R_{e}} \right )g=\frac{2}{3}\;g

Now, weight of the body at depth d,
W’ = mg’
W’ = m × 23\frac{2}{3}g  = 23\frac{2}{3} mg

W’ = 23\frac{2}{3} W

\Rightarrow W = 23\frac{2}{3} × 250  =  166.66 N

Therefore, the weight of a body at one-third of the way down to the center of the earth = 166.66 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

Sol:

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:
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=g  RE  hRE+h\\\Rightarrow v^{2}=\frac{g\;R_{E}\;h}{R_{E}+h}\\
Where, g=GMRE2\\g=\frac{GM}{R_{E}^{2}}\\ = 9.8 ms-2
Therefore, v2 (RE + h) = 2gREh
\\\Rightarrow v2RE = h (2gRE – v2)

\\\Rightarrow h=RE  v22gREv2=(6.4×106)×(4×103)22×9.8×6.4×106(4×103)2h=\frac{R_{E}\;v^{2}}{2gR_{E}-v^{2}}=\frac{(6.4\times 10^{6})\times (4\times 10^{3})^{2}}{2\times 9.8\times 6.4\times 10^{6}-(4\times 10^{3})^{2}}\\
\\\Rightarrow h = 9.356 × 105  m
Therefore, the missile reaches at height of 9.36 × 105 m from the surface.

Now, the distance from the center of the earth = h + RE = 9.36 × 105 + 6.4 x 106 = 7.33 x106 m

Therefore, the greatest height the missile can attain before falling back to the earth is   9.356 × 105 m and the final distance of the missile from the center of the earth is 7.33 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.

Sol:

Given,

Escape velocity of the Earth’s surface, vesc = 11.2 km/s
Projection
velocity of the rocket, vP = 2vesp

Let, mass of the body = m kg

And, the velocity of the rocket at a distance very far away from the surface of earth = vF

Total energy of the rocket on the surface = 12\frac{1}{2} mvp2 – 12\frac{1}{2}mv2esp

Total energy of the rocket at a distance very far away from the Earth = 12\frac{1}{2} mvF2

Now, according to the law of conservation of energy, we have:
12\frac{1}{2}mvP2 – 12\frac{1}{2}mvESC2  =  12\frac{1}{2}mvF2

mvESC2  =  12\frac{1}{2} mvF2
vF = (vp)2(vesc)2\\\sqrt{(v_{p})^{2}-(v_{esc})^{2}}\\    [Since, vP = 2vesp ]
vF = (2vesp)2(vesc)2\\\sqrt{(2v_{esp})^{2}-(v_{esc})^{2}}\\
vF =3\\\sqrt{3} vesc

\Rightarrow vF =3\sqrt{3} × 11.2 = 19.39 km/s

Therefore, the speed of rocket at a distance far away from the surface of earth is 19.39 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 × 106 m; G = 6.67 × 10–11 N m2 kg–2

Sol:

Given:

Mass of the satellite, m = 420000 kg

Radius of the Earth, RE = 6.4 × 106 m

Mass of the Earth, ME = 6.0 × 1024 kg
Universal gravitational constant, G = 6.67 × 10–11 Nm2 kg –2

Height of the satelliteh = 400 km = 0.40 ×106 m
We know that the Total energy of the satellite at height ‘h’ =12mv2+[GME  mRE+h]=\frac{1}{2}mv^{2}+\left [ \frac{-GM_{E}\;m}{R_{E}+h} \right ]

Also, orbital velocity of the satellite, v =[GMERE+h]12=\left [ \frac{GM_{E}}{R_{E}+h} \right ]^{\frac{1}{2}}

Total energy at height, h=12[GMERE+h][GME  mRE+h]=\frac{1}{2}\left [ \frac{GM_{E}}{R_{E}+h} \right ]-\left [ \frac{GM_{E}\;m}{R_{E}+h} \right ]

Therefore, TE=12[GME  mRE+h]=\frac{-1}{2}\left [ \frac{GM_{E}\;m}{R_{E}+h} \right ]
This negative sign means that the ISS is bounded to Earth. This is the bound energy of the satellite.
Now, Total Energy required to send the satellite out of its orbit = – (Bound energy)
TE=12[GME  mRE+h]\\=\frac{1}{2}\left [ \frac{GM_{E}\;m}{R_{E}+h} \right ]\\
TE=12[(6.67×1011)×(6×1024)×(4.2×105)(6.4×106)+(0.380×106)]=12×[1.681×10206.78×106]\\=\frac{1}{2}\left [ \frac{(6.67\times 10^{-11})\times (6\times 10^{24})\times (4.2\times 10^{5})}{(6.4\times 10^{6})+(0.380\times 10^{6})} \right ]=\frac{1}{2}\times \left [ \frac{1.681\times 10^{20}}{6.78\times 10^{6}} \right ]\\
\\\Rightarrow  TE = 1.2396 x 1013 J

Therefore, the amount of energy required to take ISS out from the gravitational influence of earth is 1.2396 x 1013 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 109 km, 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)

Sol:

Given:
Radius of each planet, R = 103 km = 106 m

Distance between the planet, r = 1010 km = 1013 m

Mass of each planet, M = 2 × 1031 kg

For negligible speeds, v = 0

So the total energy of two planets separated by a distance ‘r’:

TE =GMMr+12mv2\frac{-GMM}{r}+\frac{1}{2}mv^{2}

Since, v = 0;
Therefore, TE =GMMr\frac{-GMM}{r}  . . . . . . . . . . . . . . . . (1)
Now, when the planets are just about to collide:

Let, the velocity of the planets = v

The centers of the two planets are at a distance of = 2R
Total kinetic energy of the two planets = 12\frac{1}{2} Mv2 + 12\frac{1}{2} Mv2 = Mv2
Total potential energy of the two planets = GMM2R\frac{-GMM}{2R}
Therefore, Total energy of the two stars = Mv2 -GMM2R\\\frac{GMM}{2R} . . . . . . . . . . . . (2)

Now, according to the law of conservation of energy :
Mv2 – GMM2R\\\frac{GMM}{2R} = GMMr\frac{-GMM}{r}\\
v2 = GMr\\\frac{-GM}{r} + GM2R\frac{GM}{2R}\\
v2 = GM × [1r+12R]\\\left [ \frac{-1}{r}+\frac{1}{2R} \right ] \Rightarrow v2 = 6.67 × 10-11 × 2 × 1031 × [1(1013)+1(2×106)]\left [ \frac{-1}{(10^{13})}+\frac{1}{(2\times 10^{6})} \right ]
\Rightarrowv2 = 6.67 × 1014
Therefore, v = (6.67 × 1014)1/2  =  2.583 × 107m/s.

Hence, the speed at which they will collide = 2.583 × 107m/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?

Sol:

Given:

Radius of spheres, R = 0.10 m

Distance between two spheres, r = 1.0 m

Mass of each sphere, M = 100 kg

21

Two spheres

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 centers 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 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

Sol:

Given:

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

Mass of Earth, M = 6 x 1024 kg

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

Therefore, gravitational potential at height ‘h’ on the geo-stationary 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 geo-stationary 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).   

Sol.

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 = 10 × 2 × 1030 = 2 × 1031 kg
= Mass of the object
R = Radius of the star = 10 km = 1 ×104 m
Therefore, fG = (6.67×1011)×(2×1031)  m(1×104)2=1.334×1013m  N\\\frac{(6.67\times 10^{-11})\times (2\times 10^{31})\;m}{(1\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 × (104) × 4 × (3.14)2 × (10)2 = (3.9 ×107m) 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; the radius of the orbit of mars = 2.28 ×108 km; G = 6.67×10-11 N m2 kg–2

Sol:

Given,

Mass of the Sun, M = 2 × 1030 kg

Mass of the spaceshipm= 2000 kg

Radius of Mars, r = 3395 km = 3.395 × 106 m

Mass of Mars, Mm = 6.4 × 10 23 kg

Orbital radius of Mars, R = 2.28 × 10km = 2.28 × 1011 m

Universal gravitational constant, G = 6.67 × 10–11 m2 kg– 2

Now,

Potential energy of the spaceship due to the gravity of Sun = GMmsR\frac{-GMm_{s}}{R}
Potential energy of the spaceship due to gravity of Mars = GMM  msr\frac{-GM_{M}\;m_{s}}{r}

As the spaceship is stationed on Mars, its velocity is ‘zero’ and thus, its kinetic energy is also ‘zero’.
Thus, Total energy of the spaceship =GMM  msrGMmsR\\=\frac{-GM_{M}\;m_{s}}{r}-\frac{GMm_{s}}{R}
TE =Gms[MR+MMr]\\=-Gm_{s}\left [ \frac{M}{R}+\frac{M_{M}}{r} \right ]

The negative sign means that the system is inbound state.

Therefore, energy required to launch the spaceship out of the solar system:
TE = – (bound energy)
TE=Gms[MR+MMr]\\=Gm_{s}\left [ \frac{M}{R}+\frac{M_{M}}{r} \right ]

Now, on substituting the values of G, M, R, ms, MM and r we will get:

TE =(6.67×1011)×(2×103)×[2×10302.28×1011+6.4×10233.395×106]=(6.67\times 10^{-11})\times (2\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 ] \\\Rightarrow   13.34×108×[87.7×1017  +  1.885×1017]=1.195×1012J13.34\times 10^{-8}\times \left [ 87.7\times 10^{17} \;+\;1.885\times 10^{17}\right ]\boldsymbol{=1.195\times 10^{12}J}

Therefore, the total amount of energy required to launch a spaceship stationed on Mars to out of the solar system = 1.195 × 1012 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

Sol.

Given:

Mass of Venus, M = 4.8 × 1024 kg

Initial velocity of the missile, v = 2 km/s = 2 × 103 m/s

Radius of Venus, = 6052 km = 6.05 x 106 m

Universal gravitational constant, G = 6.67× 10–11 N m2 kg–2

Let the mass of the missilem kg

Since, the Initial kinetic energy of the missile = 12\frac{1}{2} mv2
and the Initial potential energy of the missile = GMmR\frac{-GMm}{R}
Thus, total initial energy =12mv2+(GMmR)=\frac{1}{2}mv^{2}+\left ( \frac{-GMm}{R} \right )

It is given that 25 % of initial kinetic energy is lost in overcoming the atmospheric resistance of Venus; this means that only 75% of the total kinetic energy is available for propelling it upwards.

Hence, the total available initial energy =[75100×12mv2]GMmR=0.375mv2GMmR=\left [ \frac{75}{100}\times \frac{1}{2}mv^{2} \right ]-\frac{GMm}{R}=0.375mv^{2}-\frac{GMm}{R}

Let ‘h’ be the maximum height attained by the missile.
Now, at height ‘h’ the final velocity = 0 and hence, the kinetic energy = 0

Therefore, the total energy of the missile at height ‘h’ = GMmR+h\frac{-GMm}{R+h}
Now, according to the law of conservation of energy:
      0.375mv2GMR=GMmR+h\\\Rightarrow\;\;\; 0.375mv^{2}-\frac{GM}{R}=\frac{-GMm}{R+h} \\\Rightarrow         0.375  v2=GMm(1R1R+h)\;\;\; 0.375\;v^{2}=GMm\left ( \frac{1}{R}-\frac{1}{R+h} \right ) 0.375v2=GM×[hR  (R+h)]\\\Rightarrow 0.375 v^{2}=GM\times \left [ \frac{h}{R\;(R+h)} \right ]\\ \\\Rightarrow   R+hh=GM0.375  v2R\frac{R+h}{h}=\frac{GM}{0.375\;v^{2}R}\\ \\\Rightarrow    Rh=GM0.375  v2R1\frac{R}{h}=\frac{GM}{0.375\;v^{2}R}-1\\ \\\Rightarrow   Rh=GM0.375  v2R0.375  v2R\frac{R}{h}=\frac{GM-0.375\;v^{2}R}{0.375\;v^{2}R}\\ \\\Rightarrow   1h=GM0.375  v2R0.375  v2R2\frac{1}{h}=\frac{GM-0.375\;v^{2}R}{0.375\;v^{2}R^{2}}\\ \\\Rightarrow   h=0.375  v2R2GM0.375  v2Rh=\frac{0.375\;v^{2}R^{2}}{GM-0.375\;v^{2}R}

Now, on substituting the values of G, M, v and R we will get:

h=0.375  ×  (3  ×  103)2  ×  (6.05  ×  106)2[(6.67  ×  1011)  ×  (4.8  ×  1024)][0.375  ×  (3  ×  103)2  ×  6.05  ×  106]\Rightarrow h=\frac{0.375\;\times \;(3\;\times \;10^{3})^{2}\;\times \;(6.05\;\times \;10^{6})^{2}}{\left [ (6.67\;\times \;10^{-11})\;\times \;(4.8\;\times \;10^{24}) \right ]-[0.375\;\times \;(3\;\times \;10^{3})^{2}\;\times \;6.05\;\times \;10^{6}]}\\ h=0.375  ×  (3  ×  103)2  ×  (6.05  ×  106)2[(6.67  ×  1011)  ×  (4.8  ×  1024)][0.375  ×  (3  ×  103)2  ×  6.05  ×  106]\\\Rightarrow h=\frac{0.375\;\times \;(3\;\times \;10^{3})^{2}\;\times \;(6.05\;\times \;10^{6})^{2}}{\left [ (6.67\;\times \;10^{-11})\;\times \;(4.8\;\times \;10^{24}) \right ]-[0.375\;\times \;(3\;\times \;10^{3})^{2}\;\times \;6.05\;\times \;10^{6}]}\\ \\\Rightarrow   1.235  ×  10203.2016  ×  10142.042  ×  1013=412023.75  m=4.12  km\frac{1.235\;\times \;10^{20}}{3.2016\;\times \;10^{14}-2.042\;\times \;10^{13}}\boldsymbol{=412023.75\;m=4.12\;km}

Therefore, the maximum height that is achieved by the missile before returning back to the surface = 4.12 km.

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.

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