NCERT Solutions for Class 11 Physics Chapter 8 - Gravitation
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8.1. 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 ?
Ans. (a) No, as the gravitational forces are not affected by the medium. The gravitational impact of neighbouring matter cannot be concealed from a body by putting it inside a hollow sphere or by other means.
(b) The answer is yes. The gravitational influence of the spaceship may become measurable or noticeable if the size of the spaceship is extremely large in accordance with Earth. It is also possible to detect changes in Earth's gravity (g).
(c) Unlike force, which is inversely proportional to the square of the distance, tidal influence is proportional to the cube of the distance. Because the moon's distance from the ocean water on Earth is so small in comparison to the distance between the Sun and the ocean water on Earth. As a result, the moon's pull has a higher tidal effect than the Sun's pull.
8.2. 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 mass of the earth/mass of the body.
(d) The formula–G Mm(1/r2 – 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.
Ans. (a) decreases
The value of acceleration due to gravity at a height 'h' above the ground gh is given $$\text{by ,} \space \text{g}_h = \text{g}\begin{bmatrix}1-\frac{2\text{h}}{\text{R}_e} \end{bmatrix}\text{where, R}_e \text{ is the radius of}$$
the Earth. Therefore, gh decreases with an increase in height.
(b) decreases Acceleration
due to gravity at depth d is given by the relation : $$\text{g(d)} = \left(1 - \frac{d}{\text{R}_e} \right)\text{g} $$ It is clear from the given relation that acceleration due to gravity decreases with increasing in depth.
(c) mass of the body
$$ \text{As g =} \frac{\text{GM}_e}{\text{R}_e^2} $$ By formula it is clear that the value of g does not depend on the mass of the falling body.
(d) more
The gravitational potential of any point at a distance from the centre of the earth is $$\text{V}{(r)} = -\frac{\text{GmM}}{\text{r}}$$ So, potential energy difference V(r2) – V(r1) $$\Delta \text{V} = - \text{ GmM}\left(\frac{1}{r_2}-\frac{1}{r_1}\right)\\ = + \text{ GmM}\left(\frac{r_2-r_1}{r_1r_2}\right) \\ = \frac{\text{ GM}}{r_1r_2}\text{m}(r_2-r_1)$$ = gm(r2 – r1)
= mgr2 – mgr1
Assuming r2 – r1 ∝ R2
$$\text{Thus, } \Delta\text{V} = -\text{GmM}\left(\frac{1}{r_2}-\frac{1}{r_1}\right)$$
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?
Ans. Here, Te = 1 year $$\text{T}_p = \frac{\text{T}_e}{2} = \frac{1}{2} \text{ years}$$re = 1 A.U.
Using Kepler's third law , we have
$$r_p = r_e\left(\frac{\text{T}_p}{\text{T}_e}\right)^{2/3} \\\Rightarrow \space \space r_p = 1\left(\frac{\text{1/2}}{\text{1}}\right)^{2/3} = 0.63 \text{ A.U.}$$
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 × 108 m. Show that the mass of Jupiter is about one-thousandth that of the sun.
Ans. For a satellite of Jupiter, orbital period,
T1 = 1.769 days
= 1.769 × 24 × 60 × 60 Sec
Radius of the orbit of satellite,
r1 = 4.22 × 108 m
Mass of Jupiter M1, is given by $$\text{M}_1 = \frac{4\pi^2 r_1^3}{\text{GT}_1^2} = \frac{4\pi^2 × (4.22 × 10^8)^3}{\text{G }× (1.769 × 24×60×60)^2} \space \space ...\text{(i)}$$
We know that the orbital period of Earth around the Sun,
T = 1 year = 365.25 × 24 × 60 × 60 s;
Orbital radius, r = 1 A.U. = 1.496 × 1011 m Mass of Sun is given by $$\text{M} = \frac{4 \pi^2\text{r}^3}{\text{GT}^2} \\ = \frac{4\pi^2 × (1.496 × 10^{11})^3}{\text{G} × (365.25 × 24 × 60×60)^2} \space \space ...\text{(ii)} \\ \text{Dividing equation (ii) by (i), we get} \\ \frac{\text{M}}{\text{M}_1} = \frac{4\pi^2 ×(1.496 × 10^{11})^3}{\text{G} × (365.25×24×60×60)^2}× \frac{\text{G} × (1.769 ×24×60×60)^2}{4\pi^2 ×(4.22×10^8)^3}= 1046 \\ \text{or } \space \space \frac{\text{M}_1}{\text{M}} = \frac{1}{1046} ≈ \frac{1}{1000} \\ \Rightarrow \space \space \text{M}_1 = \frac{1}{1000}\text{M}$$
8.5. 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.
Ans. Here, r = 50,000 ly
= 50,000 × 9.46 × 1015 m
= 4.73 × 1020 m
M = 2.5 × 1011 solar mass
= 2.5 × 1011 × (2 × 1030) kg
= 5.0 × 1041kg
We know that $${M} = \frac{4\pi^2r^3}{\text{GT}^2} \\ \text{or } \space \space \text{T} = \left( \frac{4\pi^2r^3}{\text{GM}}\right)^{1/2} \\ = \begin{bmatrix}\frac{4× (22/7)^2 × (4.73 ×10^{20})^3}{(6.67×10^{-11})×(5.0×10^{41})} \end{bmatrix}^{1/2}$$
= 1.12 × 1016 s.
= 3.55 × 108 years
8.6. 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.
Ans. (a) An orbiting satellite's total energy is negative of its kinetic energy if the potential energy zero is at infinity.
(b) The energy necessary to launch an orbiting satellite out of the gravitational influence of Earth is less than the energy required to project a stationary object of the same height (as the satellite) out of the gravitational influence of Earth.
8.7. 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?
Ans. The escape speed $$v_e =\sqrt{\frac{2\text{GM}_e}{\text{R}_e}} = \sqrt{\text{2gR}_e}.$$ Hence,
(a) No, the escape speed of a body from the Earth does not depend on the mass of the body.
(b) No, the escape speed does not depend on the location from where a body is projected.
(c) No, the escape speed does not depend on the direction of projection of a body.
(d) Yes, the escape speed of a body depends upon the height of the location from where the body is projected, because the escape velocity depends upon the gravitational potential at the point from which it is projected and this potential depends upon height also.
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.
Ans. (a) According to Kepler's second law, the comet's linear speed is changeable. When a comet approaches the Sun, its speed increases. The comet's speed is very slow when it is far from the Sun.
(b) Angular speed varies slightly as well.
(c) The angular momentum of the comet is constant.
(d) Kinetic energy does not maintain its constant value.
(e) Along the journey, potential energy varies.
(f) Total energy remains constant throughout the orbit
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.
Ans. (a) Because blood flow to the feet isn't boosted in zero gravity, that is why the astronaut doesn't suffer from swollen feet.
(b) There is more blood supply to the astronaut's face. As a result, the astronaut's face will swell.
(c) Astronauts may get headaches because of the increased blood supply to their faces.
(d) Because space has multiple orientations, an astronaut may experience orientational issues.
8.10. 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.
Ans. The potential is the same at all places inside a hollow spherical shell. As a result, gravitational intensity is zero, which is the inverse of gravitational potential gradient. The gravitational forces acting on any particle at any position inside a spherical shell will be symmetrically placed due to zero gravitational intensity. As a result, if the upper hemispherical shell is removed, the net gravitational force acting on a particle at P is downwards. The direction of gravitational intensity will be along c. Since gravitational intensity is the gravitational force per unit mass. So, option (iii) is right.
8.11. 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.
Ans. Using the explanation given in the solution of the previous problem, the direction of the gravitational field intensity at P will be along e. So, option (ii) is correct.
8.12. 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 × 1030 kg, mass of the earth = 6 × 1024 kg. Neglect the effect of other planets etc. (orbital radius = 1.5 × 1011 m).
Ans. Mass of Sun, M = 2 × 1030 kg
Mass of Earth, m = 6 × 1024 kg
Distance between Sun and Earth,
r = 1.5 × 1011 m
Let at the point P, the gravitational force on the rocket due to Earth
= gravitational force on the rocket due to Sun
Let, x = distance of the point P from the Earth $$\text{Then } \frac{\text{Gm}}{\text{x}^2} = \frac{\text{GM}}{\text{(r - x)}^2} \\ \Rightarrow \frac{\text{(r - x)}^2}{x^2} = \frac{\text{M}}{\text{m}} = \frac{2×10^{30}}{6×10^{24}} = \frac{10^6}{3} \\ \Rightarrow \frac{\text{r - x}}{\text{x}} = \frac{10^3}{\sqrt{3}} \\ \Rightarrow \frac{\text{r}}{\text{x}} = \frac{10^3}{\sqrt{3}} + 1 \approx \frac{10^3}{\sqrt{3}} \\ \Rightarrow \text{or x = } \frac{\sqrt{3}r}{10^3} \\ = \frac{1.732 × 1.5× 10^{11}}{10^3} = 2.6 ×10^8 \text{ m}.$$
8.13. 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.
Ans. The mean orbital radius of the Earth around the
Sun, R = 1.5 × 108 km = 1.5 × 1011 m
Time period of revolution
T = 365.25 × 24 × 60 × 60 s
Let the mass of the Sun be M and that of Earth be m.
According to law of gravitation
$$\text{F} = \text{G}\frac{\text{Mm}}{\text{R}^2}\space \space \text{ ...(i)}$$
Centripetal force, $$\text{F} = \frac{\text{mv}^2}{\text{R}}= \text{m.R.ω}^2\space \space \text{ ...(ii)}$$
From equation (i) and (ii), we have $$\frac{\text{GMm}}{\text{R}^2} = \text{m.R.ω}^2 \\ = \frac{\text{mR.4}\pi^2}{\text{T}^2} \begin{bmatrix}\because \text{ω} = \frac{2\pi}{\text{T}} \end{bmatrix} \\ \therefore \space \space \text{M = } \frac{4\pi^2\text{R}^3}{\text{G.T}^2} \\ = \frac{4×(3.15)^2×(1.5×10^{11})^3}{6.67×10^{-11}×(365.25×24×60×60)^2}$$ = 2.009 × 1030 kg = 2.0 × 1030 kg
8.14. A saturn year is 29.5 times the earth year. How far is the saturn from the sun if the earth is 1.50 × 108 km away from the sun ?
Ans. By Kepler‘s Law
T2 ∝ R3 $$\therefore \space \space \frac{\text{T}_s^2}{\text{T}_e^2}= \frac{\text{R}_s^3}{\text{R}_e^3}$$ where, subscripts s and e refer to the Saturn and Earth respectively. $$\text{Now }\space\space \frac{\text{T}_s}{\text{T}_e} = 29.5 \text{ (given)}; \\ \text{R}_e = 1.50 × 10^8 \text{ km} \\ \left(\frac{\text{R}_s}{\text{R}_e} \right)^3 = \left(\frac{\text{T}_s}{\text{T}_e} \right)^2$$
Rs = Re × [(29.5)2]1/3
= 1.50 × 108 × (870.25)1/3 km
= 1.43 × 109 km
= 1.43 × 1012 m
∴ Distance of Saturn from sun
= 1.43 × 1012 m
8.15. 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 ?
Ans. Let gh be the acceleration due to gravity at a height equal to half the radius of the Earth
$$\text{(h = }\frac{\text{R}}{2})$$ and g its value on Earth’s surface. Let the body have mass m.
We know that
$$\frac{\text{g}_h}{\text{g}} = \left(\frac{\text{R}}{\text{R+h}} \right)^2 \\ \text{or }\space \frac{\text{g}_h}{\text{g}} = \left(\frac{\text{R}}{\text{R}+\frac{\text{R}}{2}} \right)^2 \\ = \left(\frac{2}{3} \right)^2 = \frac{4}{9}$$ Let W be the weight of body on the surface of Earth and Wh the weight of the body at height h. $$\text{Then, } \frac{\text{W}_h}{\text{W}} = \frac{\text{mg}_h}{\text{mg}} = \frac{\text{g}_h}{\text{g}} =\frac{4}{9} \\ \text{or } \space \space \text{W}_h =\frac{4}{9}\text{ W = }\frac{4}{9}× 63\text{ N = 28 N} $$
8.16. Assuming the earth to be a sphere of uniform mass density, how much would a body weigh half way down to the centre of the earth if it weighed 250 N on the surface ?
Ans. By formula,
$$\text{g}_d = \text{g}\left(1-\frac{\text{d}}{\text{R}}\right) \\ \Rightarrow \text{mg}_d = \text{mg}\left(1-\frac{\text{d}}{\text{R}}\right) \\ \text{Here, } \text{d} = \frac{\text{R}}{2} \\ \therefore \space \space \text{mg}_d = (250) × \left(1-\frac{\text{R}/2}{\text{R}} \right)\\ = 250 × \frac{1}{2} = 125 \text{ N.}$$
8.17. 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.
Ans. Initial kinetic energy of rocket $$= \frac{1}{2}\space \text{mv}^2 = \frac{1}{2}× \text{m}× (5000)^2$$ = 1.25 × 107 mJ
At distance r from centre of earth, kinetic energy becomes zero
∴ Change in kinetic energy
= 1.25 × 107 m – 0 = 1.25 × 107 m J
This energy i.e., kinetic energy changes into potential energy.
Initial potential energy at the surface of earth $$= - \frac{\text{GM}_em}{r}\\ = \frac{-(6.67 × 10^{11}) × (6 × 10^{24})m}{6.4 × 10^6} $$ = – 6.25 m × 107J
Final potential energy at distance, $$ r = - \frac{\text{GM}_em}{r}\\ = \frac{-(6.67 × 10^{-11}) × (6 × 10^{24})m}{r}\\ = -4 × 10^{14} \frac{\text{m}}{\text{r}}\text{ J}$$ ∴ Change in potential energy $$= 6.25 × 10^7 \text{ m – 4} × 10^{14} \frac{ \text{m}}{\text{r}}$$ Using law of conservation of energy, $$6.25 × 10^7 \text{m} - \frac{4×10^{14}m}{r} = 1.25 × 10^7 \text{m} \\ \text{i.e., }\space \space r = \frac{4 × 10^{-14}}{5 × 10^7}\text{m}$$ = 8 × 107m.
8.18. 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.
Ans. Let ves be the escape speed from surface of Earth having a value
ves = 11.2 kg s–1
= 11.2 × 103 m s–1.
By definition,
$$\frac{1}{2}\text{mv}^2_e = \frac{\text{GMm}}{\text{R}^2}$$ When a body is projected with a speed
vi = 3ves = 3 × 11.2 × 103 m/s,
The it will have a final speed, vf such that
$$\frac{1}{2}\text{mv}_f^2 = \frac{1}{2}\text{mv}_i^2 - \frac{\text{GMm}}{\text{R}^2} \\ = \frac{1}{2}\text{mv}_i^2 - \frac{1}{2}\text{mv}_{es}^2 \\ \Rightarrow \space \space v_f = {\sqrt{v_i^2 - v_{es}^2}} \\ = \sqrt{(3× 11.2×10^3) - (11.2×10^3)^2} \\ = 11.2 × 10^3 × \sqrt{8} \\ = 31.7 × 10^3 \text{ ms}^{–1} \text{ or } 31.7 \text{ km s}^{–1}$$
8.19. 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.
Ans. Total energy of orbiting satellite at a height
h = P.E. + K.E. $$= -\frac{\text{GMm}}{\text{(R+h)}} + \frac{1}{2}\text{mv}^2 \\ = -\frac{\text{GMm}}{\text{(R+h)}} + \frac{1}{2}\text{m}\frac{\text{GM}}{\text{(R + h)}} \\ = -\frac{\text{GM}}{\text{2(R + h)}}$$ Energy expended to rocket the satellite out of the earth‘s gravitational influence
= – (total energy of the orbiting satellite)
$$ = \frac{\text{GM}}{\text{2(R + h)}} \\ = \frac{(6.67 ×10^{-11})×(6×10^{24})×200}{2× (6.4×10^6 + 4 × 10^5)} \\ = 5.9 × 10^9 \text{J}$$
8.20. 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).
Ans. Here, mass of each star,
M = 2 × 1030 kg
Initial potential between two stars,
r = 109 km = 1012 m.
Initial potential energy of the system $$= - \frac{\text{GMm}}{r} $$ Total K.E. of the stars $$= \frac{1}{2}\text{Mv}^2 + \frac{1}{2}\text{Mv}^2$$
where, v is the speed of stars with which they collide. When the stars are about to collide, the distance between their centres,
r′ = 2R.
∴ Final potential energy of two stars $$= -\frac{\text{GMM}}{\text{2R}}$$ Since gain in K.E. is at the cost of loss in P.E. $$\therefore \space \space \text{Mv}^2 = -\frac{\text{GMM}}{r}-\left(-\frac{\text{GMM}}{2\text{R}} \right) \\ = -\frac{\text{GMM}}{r}+\frac{\text{GMM}}{\text{2R}} \\ \text{or }\space \space 2×10^{30} v^2 = - \frac{6.67×10^{-11}×(2×10^{30})^2}{10^{12}}+\frac{6.67×10^{-11}×(2×10^{30})^2}{2×10^7} \\ = - 2.668 × 10^{38} +1.334 × 10^{43}\\ = 1.334 × 10^{43} \text{J} \\ \therefore\space \space v = \sqrt{\frac{1.334 × 10^{43}}{2× 10^{30}}}\\ = 2.583 × 10^6 \text{ms}^{-1}.$$
8.21. 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 mid-point 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 ?
Ans. Here G = 6.67 × 10-11 Nm2 kg-2;
M = 100 kg;
R = 0.1 m,
Distance between the two spheres,
d = 1.0 m
Suppose that the distance of either sphere from the mid-point of the line joining their centre is r.
Then r = d/2 = 0.5 m.
The gravitational field at the mid-point due to two spheres will be equal and opposite.
Hence, the resultant gravitational field at the mid point = 0
Gravitational potential at the mid-point $$= \left(-\frac{\text{GM}}{\text{r}} \right)×2 \\ = -\frac{6.67×10^{-11}×100×2}{0.5}\\ = - 2.668 × 10^{-8} \text{J kg}^{-1}.$$ The object placed at that point would be in stable equilibrium.
Therefore, the gravitational potential and force at the mid-point of the line connecting the centers of the two spheres is
= – 2.668 × 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.
8.22. 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.
Ans. Distance of satellite from the center of earth
R = r + x = 6400 + 36000
= 42400 km = 4.24 × 107 m
Using potential, $$\text{V}= -\frac{\text{GM}}{\text{R}}, \text{we get} \\ \text{V = -}\frac{(6.67×10^{-11})×(6×10^{24})}{(4.24×10^7)} \\ = -9.44 × 10^6\text{J kg}^{-1} $$
Therefore, the gravitational potential due to Earth‘s gravity on a geostationary satellite orbitating earth is – 9.439 × 106 J/Kg.
8.23. 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).
Ans. Acceleration due to gravity of the star, $$\text{g} = \frac{\text{GM}}{\text{R}^2} ...\text{(i)}$$ Here, M is the mass and R is the radius of the star.
The outward centrifugal force acting on a body of mass m at the equator of the star $$= \frac{\text{mv}^2}{\text{R}} = \text{mRω}^2 \space \space \text{...(ii)}$$ From equation (i), the acceleration due to the gravity of the star $$\text{g} = \frac{6.67×10^{-11}×2.5×2×10^{30}}{(12×10^3)^2}$$ = 2.316 × 1012 m/s2
∴ Inward force due to gravity on a body of mass m
= m × 2.316 × 1012 N
From equation (ii), the outward centrifugal force
= mRω2 $$= m × (12× 10^3) × \left( \frac{2\pi ×1.5}{-1}\right)^2$$ = m × 1.06 × 106 N
Since, the inward force due to gravity on a body at the equator of the star is about 2.2 million times more than the outward centrifugal force, the body will remain stuck to the surface of the star.
8.24. 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 × 108 km; G = 6.67 × 10-11 N m2 kg–2.
Ans. Let R be the radius of the orbit of Mars and R’ be the radius of Mars.
M be the mass of the Sun and M’ be the mass of Mars.
If m is the mass of the spaceship, then Potential energy of space-ship due to gravitational attraction of the Sun $$= - \frac{\text{GMm}}{\text{R}}$$Potential energy of space-ship due to gravitational attraction of Mars$$= - \frac{\text{GM}'\text{m}}{\text{R}'}$$ Since the K.E. of space ship is zero, therefore,
Total energy of spaceship $$= - \frac{\text{GMm}}{\text{R}} - \frac{\text{GM}'\text{m}}{\text{R}'} \\ = -\text{Gm}\left(\frac{\text{M}}{\text{R}}+\frac{\text{M}'}{\text{R}'} \right)$$ ∴ Energy required to rocket out the spaceship from the solar system
= – (total energy of spaceship) $$= -\begin{bmatrix}-\text{Gm} \left(\frac{\text{M}}{\text{R}}+\frac{\text{M}'}{\text{R}'} \right)\end{bmatrix} \\ = -\text{Gm}\begin{bmatrix}\frac{\text{M}}{\text{R}}+\frac{\text{M}'}{\text{R}'}\end{bmatrix} \\ = 6.67 × 10^{-11} × 1000 × \begin{bmatrix}\frac{2× 10^{30}}{2.28× 10^{11}} + \frac{6.4 × 10^{23}}{3395 × 10^3} \end{bmatrix} \\ = 6.67 × 10^{-8 }\begin{bmatrix} \frac{20}{2.28} + \frac{6.4}{33.95}\end{bmatrix} × 10^{18}\text{J} \\ = 5.98 × 10^{11}\text{J}.$$
8.25. 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.
Ans. Initial kinetic energy of rocket $$= \frac{1}{2}\text{mv}^2;$$ Initial P.E. of rocket $$= - \frac{\text{GMm}}{\text{R}}$$
where m = Mass of rocket,
M = Mass of Mars,
R = Radius of Mars
∴ Total Initial energy
$$=\frac{1}{2} \text{mv}^2 - \frac{\text{GMm}}{\text{R}}$$ Since
20% of K.E. is lost, only 80% remains to reach the height.
∴ Total initial energy available $$=\frac{4}{5}×\frac{1}{2} \text{ mv}^2 - \frac{\text{GMm}}{\text{R}} \\ = 0.4\text{ mv}^2 - \frac{\text{GMm}}{\text{R}}$$ When
the rocket reaches the highest point, at a height h above the surface, its K.E. is zero and P.E. $$= - \frac{\text{GMm}}{\text{R + h}}$$ Using principle of conservation of energy. $$0.4 \text{ mv}^2 - \frac{\text{GMm}}{\text{R}} = -\frac{\text{GMm}}{\text{R + h}} \\ \Rightarrow\space \space \frac{\text{GMm}}{\text{R + h}}= - \frac{\text{GMm}}{\text{R}} - 0.4 \space \text{mv}^2 \\ \\ \Rightarrow\space \space \frac{\text{GM}}{\text{R + h}}= \frac{\text{GM}}{\text{R}} - 0.4 \space \text{v}^2 \\ \Rightarrow \space \space \frac{\text{GM}}{\text{R+h}} = \frac{1}{\text{R}}[\text{GM - 0.4 \text{Rv}}^2] \\ \Rightarrow \frac{\text{R + h}}{\text{R}} = \frac{\text{GM}}{\text{MG}-0.4\text{Rv}^2} \\ \Rightarrow \frac{\text{h}}{\text{R}} = \frac{\text{GM}}{\text{MG}-0.4\text{Rv}^2}-1 \\ \Rightarrow \frac{\text{h}}{\text{R}} = \frac{\text{0.4 Rv}^2}{\text{GM}-0.4\text{Rv}^2} \\\Rightarrow \space \space \text{h} = \frac{\text{0.4 R}^2\text{v}^2}{\text{GM}-0.4\text{Rv}^2} \\ \\ \Rightarrow \space \space h = \frac{0.4× (2×10^3)^2×(3.395×10^6)^2}{6.67×10^{-11} × 6.4× 10^{23}-0.4} \text{m} × (2×10^3)^2 × 3.395 × 10^6 \approx 495 \text{km.}$$
