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 spaceship 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 the medium, and it is the virtue of each and every matter. So the shield would exert gravitational forces.
(b). Yes, if the spaceship is large enough, then the astronaut will definitely detect Mars’s 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 the 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
The radius of Earth’s orbit, RE = 1 AU
Thus, the time taken by the planet to complete one complete revolution
TP =
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 = 4Ï€2RI03 / GTI02 . . . . . . . . . . . . . . . (1)
Where;
MJ = Mass of Jupiter
G = Universal gravitational constant
Also,
The orbital period of the earth
TE = 365.25 days = 365.25 × 24 × 60 × 60 s
The orbital radius of the earth, RE = 1 AU = 1.496 × 1011 m
We know that the mass of the sun is
MS =
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=5×104 ly
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=(4Ï€2r3/GM)1/2
=
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 the 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 the 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
(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 the 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
Solution: (iii) c
Reason:
Inside a hollow sphere, gravitational forces on any particle at any point are 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, and the 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 is due to Earth.
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 =
Q. 14. A Saturn year is 29.5 times the Earth year. How far is 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,
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’ =
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 =
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:
Mass of the satellite, m = 200 kg
Radius of the Earth, Re = 6.4 × 106 m
= (1/2)mv² + [ -GmMe / (Re + h) ]
Total K.E. of the stars = 1/2Mv2 + 1/2Mv2 = Mv2
Therefore, the final potential energy of two stars when they are close to each other = -GMm/2R
Solution:
Given:
Radius of spheres, R = 0.10 m
Distance between two spheres, r = 1.0 m
Mass of each sphere, M = 100 kg
From the above figure, ‘A’ is the mid-point 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
Q.22: As you have learnt in the text, a geostationary satellite orbits the earth at the height of nearly 36,000 km from the surface of the earth. What is the potential due to the 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 the 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 =
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 =
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? The mass of the spaceship = 1000 kg; the mass of the sun = 2×1030 kg; the mass of Mars = 6.4×1023 kg; the radius of Mars = 3395 km; the radius of the orbit of Mars = 2.28 ×108 km; G = 6.67×10-11 N m2kg–2.
Solution:
The mass of the spaceship, ms =1000 kg
the mass of the Sun, M=2×1030kg
the mass of Mars, mm =6.4×1023kg
The radius of the orbit of Mars, R=2.28×1011 m
The 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
The 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.
The 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)
=
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? The mass of mars = 6.4×1023 kg; the 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
The mass of the rocket = m
The mass of Mars ( M) = 6.4 × 10²³ Kg
The radius of Mars (R) = 3395 km = 3.395 × 106 m
The initial Potential energy = – GMm/R
The 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 the 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)
.png)
0 Comments