NCERT 11th Physics Chapter 8, class 11 Physics chapter 8 solutions<\/p>\n\n\n\n
Page No: 201<\/p>\n\n\n\n
Excercises<\/h2>\n\n\n\n
8.1. Answer the following:<\/strong> Answer<\/strong><\/p>\n\n\n\n (a) We cannot shiwld a body from the gravitational influence of nearby matter, because the gravitational force on a body due to nearby matter is independent of the presence of other matter, whereas it is not so in case of electric forces. It means the gravitational screens are not possible.<\/p>\n\n\n\n (b) Yes, if the size of the space station is large enough, then the astronaut will detect the change in Earth\u2019s gravity (g<\/em>). 8.2. Choose the correct alternative:<\/strong> Answer<\/strong><\/p>\n\n\n\n (a) decreases 8.3. 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?<\/strong><\/p>\n\n\n\n Answer<\/strong><\/p>\n\n\n\n Time taken by the Earth to complete one revolution around the Sun, 8.4. Io, one of the satellites of Jupiter, has an orbital period of 1.769 days and the radius of the orbit is 4.22 \u00d7 108<\/sup> m. Show that the mass of Jupiter is about one-thousandth that of the sun.<\/strong><\/p>\n\n\n\n Answer<\/strong><\/p>\n\n\n\n Orbital period of I0<\/sub> , T<\/em>I0<\/sub> = 1.769 days = 1.769 \u00d7 24 \u00d7 60 \u00d7 60 s<\/p>\n\n\n\n Orbital radius of I0<\/sub> , R<\/em>I<\/sub>0<\/sub> = 4.22 \u00d7 108<\/sup> m<\/p>\n\n\n\n Satellite I0<\/sub> is revolving around the Jupiter<\/p>\n\n\n\n Mass of the latter is given by the relation:<\/p>\n\n\n\n M<\/em>J<\/sub> = 4\u03c02<\/sup>R<\/em>I0<\/sub>3<\/sup> \/ GT<\/em>I0<\/sub>2<\/sup> …..(i)<\/strong><\/p>\n\n\n\n Where,<\/p>\n\n\n\n M<\/em>J<\/sub> = Mass of Jupiter<\/p>\n\n\n\n G = Universal gravitational constant<\/p>\n\n\n\n Orbital period of the earth,<\/p>\n\n\n\n Te <\/sub>= 365.25 days = 365.25 \u00d7 24 \u00d7 60 \u00d7 60 s<\/p>\n\n\n\n Orbital radius of the Earth,<\/p>\n\n\n\n R<\/em>e <\/sub>= 1 AU = 1.496 \u00d7 1011 <\/sup>m<\/p>\n\n\n\n Mass of sun is given as:<\/p>\n\n\n\n M<\/em>s<\/sub> = 4\u03c02<\/sup>R<\/em>e<\/sub>3<\/sup> \/ GT<\/em>e<\/sub>2 <\/sup> ……(ii)<\/strong><\/p>\n\n\n\n \u2234 M<\/em>s<\/sub> \/ M<\/em>J<\/sub> = (4\u03c02<\/sup>R<\/em>e<\/sub>3<\/sup> \/ GT<\/em>e<\/sub>2<\/sup>) \u00d7 (GT<\/em>I0<\/sub>2<\/sup> \/ 4\u03c02<\/sup>R<\/em>I0<\/sub>3<\/sup>) = (R<\/em>e<\/sub>3<\/sup> \u00d7 T<\/em>I0<\/sub>2<\/sup>) \/ (R<\/em>I0<\/sub>3<\/sup> \u00d7 T<\/em>e<\/sub>2<\/sup>)<\/p>\n\n\n\n Substituting the values, we get:<\/p>\n\n\n\n = (1.769 \u00d7 24 \u00d7 60 \u00d7 60 \/ 365.25 \u00d7 24 \u00d7 60 \u00d7 60)2<\/sup> \u00d7 (1.496 \u00d7 1011<\/sup> \/ 4.22 \u00d7 108<\/sup>)3<\/sup><\/p>\n\n\n\n = 1045.04<\/p>\n\n\n\n \u2234 M<\/em>s<\/sub> \/ M<\/em>J<\/sub> ~ 1000<\/p>\n\n\n\n M<\/em>s<\/sub> ~ 1000 \u00d7 M<\/em>J<\/sub><\/p>\n\n\n\n Hence, it can be inferred that the mass of Jupiter is about one-thousandth that of the Sun.<\/p>\n\n\n\n NCERT 11th Physics Chapter 8, class 11 Physics chapter 8 solutions<\/p>\n\n\n\n 8.5. Let us assume that our galaxy consists of 2.5 \u00d7 1011<\/sup> 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<\/sup> ly.<\/strong><\/p>\n\n\n\n Answer<\/strong><\/p>\n\n\n\n Mass of our galaxy Milky Way, M<\/em> = 2.5 \u00d7 1011<\/sup> solar mass 8.6. Choose the correct alternative:<\/strong> Answer<\/strong><\/p>\n\n\n\n (a) Kinetic energy 8.7. Does the escape speed of a body from the earth depend on<\/strong> Answer<\/strong><\/p>\n\n\n\n The escape velocity is indpendent of the mass of the body and the direction of projection. It depends upon the gravitational potential at the point from where the body is launched. Since, this potential depends slightly on the latitude and height of the point, therefore, the escape velocity depends slightly on these factors.<\/p>\n\n\n\n NCERT 11th Physics Chapter 8, class 11 Physics chapter 8 solutions<\/p>\n\n\n\n 8.8. 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.<\/strong><\/p>\n\n\n\n Answer<\/strong><\/p>\n\n\n\n A comet while going on elliptical orbit around the Sun has constant angular momentum and totaal energy at all locations but other quantities vary with locations.<\/p>\n\n\n\n 8.9. Which of the following symptoms is likely to afflict an astronaut in space (a) swollen feet, (b) swollen face, (c) headache, (d) orientational problem?<\/strong><\/p>\n\n\n\n Answer<\/strong><\/p>\n\n\n\n (a) Legs hold the entire mass of a body in standing position due to gravitational pull. In space, an astronaut feels weightlessness because of the absence of gravity. Therefore, swollen feet of an astronaut do not affect him\/her in space.<\/p>\n\n\n\n (b) A swollen face is caused generally because of apparent weightlessness in space. Sense organs such as eyes, ears nose, and mouth constitute a person\u2019s face. This symptom can affect an astronaut in space.<\/p>\n\n\n\n (c) Headaches are caused because of mental strain. It can affect the working of an astronaut in space.<\/p>\n\n\n\n (d) Space has different orientations. Therefore, orientational problem can affect an astronaut in space.<\/p>\n\n\n\n 8.10. 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) O.<\/strong><\/p>\n\n\n\n Answer<\/strong><\/p>\n\n\n\n Gravitational potential (V<\/em>) is constant at all points in a spherical shell. Hence, the gravitational potential gradient (dV<\/em>\/dR<\/em>) is zero everywhere inside the spherical shell. The gravitational potential gradient is equal to the negative of gravitational intensity. Hence, intensity is also zero at all points inside the spherical shell. This indicates that gravitational forces acting at a point in a spherical shell are symmetric. Since gravitational intensity at a point is defined as the gravitational force per unit mass at that point, it will also act in the downward direction. Thus, the gravitational intensity at centre O of the given hemispherical shell has the direction as indicated by arrow c<\/strong>.<\/p>\n\n\n\n 8.11. Choose the correct answer from among the given ones:<\/strong> Answer<\/strong><\/p>\n\n\n\n Gravitational potential (V<\/em>) is constant at all points in a spherical shell. Hence, the gravitational potential gradient (dV<\/em>\/dR<\/em>) is zero everywhere inside the spherical shell. The gravitational potential gradient is equal to the negative of gravitational intensity. Hence, intensity is also zero at all points inside the spherical shell. This indicates that gravitational forces acting at a point in a spherical shell are symmetric. Since gravitational intensity at a point is defined as the gravitational force per unit mass at that point, it will also act in the downward direction. Thus, the gravitational intensity at an arbitrary point P of the hemispherical shell has the direction as indicated by arrow e<\/strong>.<\/p>\n\n\n\n Hence, the correct answer is (ii).<\/p>\n\n\n\n NCERT 11th Physics Chapter 8, class 11 Physics chapter 8 solutions<\/p>\n\n\n\n 8.12. A rocket is fired from the earth towards the sun. At what distance from the earth\u2019s centre is the gravitational force on the rocket zero? Mass of the sun = 2 \u00d71030<\/sup> kg, mass of the earth = 6 \u00d7 1024<\/sup> kg. Neglect the effect of other planets etc. (orbital radius = 1.5 \u00d7 1011<\/sup> m).<\/strong><\/p>\n\n\n\n Answer<\/strong><\/p>\n\n\n\n Mass of the Sun, M<\/em>s<\/sub> = 2 \u00d7 1030<\/sup> kg<\/p>\n\n\n\n Mass of the Earth, M<\/em>
(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\u2019s pull is greater than the moon\u2019s pull. (You can check this yourself using the data available in the succeeding exercises). However, the tidal effect of the moon\u2019s pull is greater than the tidal effect of sun. Why?<\/strong><\/p>\n\n\n\n
(c) Tidal effect depends inversely upon the cube of the distance while, gravitational force depends inversely on the square of the distance. Since the distance between the Moon and the Earth is smaller than the distance between the Sun and the Earth, the tidal effect of the Moon\u2019s pull is greater than the tidal effect of the Sun\u2019s pull.<\/p>\n\n\n\n
(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 \u2013G Mm<\/em>(1\/r<\/em>2<\/sub>\u2013 1\/r<\/em>1<\/sub>) is more\/less accurate than the formula mg<\/em>(r<\/em>2<\/sub>\u2013 r<\/em>1<\/em><\/sub>) for the difference of potential energy between two points r<\/em>2<\/sub>and r<\/em>1<\/sub>distance away from the centre of the earth.<\/strong><\/p>\n\n\n\n
(b) decreases
(c) mass of the body
(d) more.<\/p>\n\n\n\n
T<\/em>e<\/sub> = 1 year
Orbital radius of the Earth in its orbit, R<\/em>e <\/sub>= 1 AU
Time taken by the planet to complete one revolution around the Sun, T<\/em>P<\/sub> = \u00bdT<\/em>e<\/sub> = \u00bd year
Orbital radius of the planet = R<\/em>p<\/sub>
From Kepler\u2019s third law of planetary motion, we can write:
(R<\/em>p<\/sub> \/ R<\/em>e<\/sub>)3<\/sup> = (T<\/em>p<\/sub> \/ T<\/em>e<\/sub>)2<\/sup>
(R<\/em>p<\/sub> \/ R<\/em>e<\/sub>) = (T<\/em>p<\/sub> \/ T<\/em>e<\/sub>)2\/3<\/sup>
= (\u00bd \/ 1)2\/3<\/sup> = 0.52\/3<\/sup> = 0.63
Hence, the orbital radius of the planet will be 0.63 times smaller than that of the Earth.<\/p>\n\n\n\n
Solar mass = Mass of Sun = 2.0 \u00d7 1036<\/sup> kg
Mass of our galaxy, M<\/em> = 2.5 \u00d7 1011<\/sup> \u00d7 2 \u00d7 1036<\/sup> = 5 \u00d7 1041<\/sup> kg
Diameter of Milky Way, d<\/em> = 105<\/sup> ly
Radius of Milky Way, r<\/em> = 5 \u00d7 104<\/sup> ly
1 ly = 9.46 \u00d7 1015<\/sup> m
\u2234r<\/em> = 5 \u00d7 104<\/sup> \u00d7 9.46 \u00d7 1015<\/sup>
= 4.73 \u00d71020<\/sup> m
Since a star revolves around the galactic centre of the Milky Way, its time period is given by the relation:
T <\/em>= ( 4\u03c02<\/sup>r3<\/sup> \/ GM<\/em>)1\/2<\/sup>
= [ (4 \u00d7 3.142<\/sup> \u00d7 4.733<\/sup> \u00d7 1060<\/sup>) \/ (6.67 \u00d7 10-11<\/sup> \u00d7 5 \u00d7 1041<\/sup>) ]1\/2<\/sup>
= (39.48 \u00d7 105.82 \u00d7 1030<\/sup> \/ 33.35 )1\/2<\/sup>
= 1.12 \u00d7 1016<\/sup> s
1 year = 365 \u00d7 324 \u00d7 60 \u00d7 60 s
1s = 1 \/ (365 \u00d7 324 \u00d7 60 \u00d7 60) years
\u2234 1.12 \u00d7 1016<\/sup> s = 1.12 \u00d7 1016<\/sup> \/ (365 \u00d7 24 \u00d7 60 \u00d7 60) = 3.55 \u00d7 108<\/sup> years.<\/p>\n\n\n\n
(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\u2019s gravitational influence is more\/less than the energy required to project a stationary object at the same height (as the satellite) out of earth\u2019s influence.<\/strong><\/p>\n\n\n\n
(b) Less<\/p>\n\n\n\n
(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?<\/strong><\/p>\n\n\n\n![\"NCERT](\"https:\/\/www.indcareer.com\/schools\/wp-content\/uploads\/2021\/03\/fig-8.12-chapter-8-class-11th.png\")
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If the upper half of a spherical shell is cut out (as shown in the given figure), then the net gravitational force acting on a particle located at centre O will be in the downward direction.<\/p>\n\n\n\n![\"NCERT](\"https:\/\/www.indcareer.com\/schools\/wp-content\/uploads\/2021\/03\/fig-1-chapter-8-class-11th.png\")
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For the problem 8.10, the direction of the gravitational intensity at an arbitrary point P is indicated by the arrow (i) d, (ii) e, (iii) f, (iv) g.<\/strong><\/p>\n\n\n\n
If the upper half of a spherical shell is cut out (as shown in the given figure), then the net gravitational force acting on a particle at an arbitrary point P will be in the downward direction.<\/p>\n\n\n\n![\"NCERT](\"https:\/\/www.indcareer.com\/schools\/wp-content\/uploads\/2021\/03\/fig-2-chapter-8-class-11th.png\")
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