Question 1

(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?

Accepted Answer

(a) No, you cannot shield a body from gravitational influence by placing it inside a hollow sphere or by any other means. This is because gravity is always attractive and acts through mass, and there is no known material or configuration that can block gravitational fields. Unlike electrical forces which can be shielded by conductors due to the presence of positive and negative charges, gravity has only one type of 'charge' (mass), so it cannot be shielded.

(b) An astronaut inside a small spaceship orbiting the Earth experiences weightlessness because both the astronaut and the spaceship are in free fall towards the Earth, creating a microgravity environment. If the space station is very large, tidal effects (variations in gravitational field strength across the station) become detectable. Hence, the astronaut can detect gravity through these tidal forces in a large space station.

(c) Although the Sun exerts a stronger gravitational force on the Earth than the Moon does, the tidal effect depends on the gradient (difference) of gravitational force across the Earth. Since the Moon is much closer to the Earth than the Sun, the difference in gravitational pull across the Earth due to the Moon is greater than that due to the Sun. Therefore, the Moon's tidal effect is greater than the Sun's. — The gravitational force cannot be shielded because gravity is always attractive and acts through mass without positive or negative charges. Weightlessness in orbit is due to free fall, and tidal effects arise from differences in gravitational pull across an object. The Moon's proximity causes a larger tidal effect despite its smaller gravitational force compared to the Sun.

Question 2

(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  $-G M m (1 / r_{2} - 1 / r_{1})$  is more/less accurate than the formula  $m g (r_{2} - r_{1})$  for the difference of potential energy between two points  $r_{2}$  and  $r_{1}$  distance away from the centre of the earth.

Accepted Answer

(a) Acceleration due to gravity decreases with increasing altitude because gravitational force decreases with distance from the Earth's center.

(b) Acceleration due to gravity decreases with increasing depth inside the Earth (assuming uniform density) because the effective mass attracting the object decreases linearly with depth.

(c) Acceleration due to gravity is independent of the mass of the body; it depends on the mass of the Earth and the distance from its center.

(d) The formula $-G M m (1 / r_{2} - 1 / r_{1})$ is more accurate than $m g (r_{2} - r_{1})$ for calculating the difference in potential energy between two points at distances $r_1$ and $r_2$ from the Earth's center, because the latter assumes constant g, which is only an approximation valid for small height differences. — Gravity decreases with altitude because gravitational force varies inversely with the square of distance. Inside Earth, gravity decreases linearly with depth due to the shell theorem. Acceleration due to gravity is independent of the object's mass. The exact potential energy difference formula accounts for variation of g with distance, making it more accurate.

Question 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?

Accepted Answer

Let the orbital period of Earth be T and orbital radius be R.

Given: New planet's period T' = T/2.

From Kepler's third law: T^2 ∝ R^3

Therefore, (T')^2 / T^2 = (R')^3 / R^3

=> (T/2)^2 / T^2 = (R')^3 / R^3

=> (1/4) = (R')^3 / R^3

=> (R')^3 = R^3 / 4

=> R' = R / (4)^{1/3} = R / 1.5874 ≈ 0.63 R

So, the orbital radius of the planet is approximately 0.63 times that of the Earth. — Using Kepler's third law, the square of the orbital period is proportional to the cube of the orbital radius. Halving the period reduces the radius by the cube root of 1/4.

Question 4

Io, one of the satellites of Jupiter, has an orbital period of 1.769 days and the radius of the orbit is  $4.22 	imes 10^{8} \mathrm{~m}$ . Show that the mass of Jupiter is about one-thousandth that of the sun.

Accepted Answer

Given:
Orbital period, T = 1.769 days = 1.769 × 24 × 3600 = 152,793.6 s
Orbital radius, r = 4.22 × 10^8 m

Using the formula for orbital period:

T^2 = \frac{4 \pi^2 r^3}{G M}

Rearranged to find mass M:

M = \frac{4 \pi^2 r^3}{G T^2}

Substitute values:

G = 6.67 × 10^{-11} N·m^2/kg^2

M = \frac{4 \pi^2 (4.22 	imes 10^8)^3}{6.67 	imes 10^{-11} (152,793.6)^2}

Calculate numerator:
(4.22 × 10^8)^3 = 7.52 × 10^{25} m^3

4 π^2 × 7.52 × 10^{25} ≈ 4 × 9.87 × 7.52 × 10^{25} ≈ 296.5 × 10^{25} = 2.965 × 10^{27}

Calculate denominator:
6.67 × 10^{-11} × (1.527936 × 10^5)^2 = 6.67 × 10^{-11} × 2.334 × 10^{10} = 1.556 × 10^0 = 1.556

Therefore,
M = 2.965 × 10^{27} / 1.556 ≈ 1.9 × 10^{27} kg

Mass of Sun = 2 × 10^{30} kg

Ratio = (1.9 × 10^{27}) / (2 × 10^{30}) = 9.5 × 10^{-4} ≈ 1/1000

Hence, the mass of Jupiter is about one-thousandth that of the Sun. — Using the orbital period formula derived from Newton's law of gravitation and centripetal force, we calculate Jupiter's mass from Io's orbital parameters and compare it with the Sun's mass.

Question 5

Let us assume that our galaxy consists of  $2.5 	imes 10^{11}$  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  $10^{5}$  ly.

Accepted Answer

Given:
Number of stars, N = 2.5 × 10^{11}
Mass of each star = 1 solar mass = 2 × 10^{30} kg
Distance of star from galactic center, r = 50,000 ly
Diameter of Milky Way = 10^{5} ly

Assuming mass inside radius r is M = N × mass of star (approximate, assuming uniform distribution)

But since r = 50,000 ly = half the diameter, mass inside radius r is about half total mass:

M = (N/2) × 2 × 10^{30} = 1.25 × 10^{11} × 2 × 10^{30} = 2.5 × 10^{41} kg

Convert r to meters:
1 ly = 9.46 × 10^{15} m
r = 50,000 × 9.46 × 10^{15} = 4.73 × 10^{20} m

Using orbital velocity formula:

v = \sqrt{\frac{G M}{r}}

Orbital period T = \frac{2 \pi r}{v} = 2 \pi r / \sqrt{G M / r} = 2 \pi \sqrt{\frac{r^3}{G M}}

Calculate T:

G = 6.67 × 10^{-11} N·m^2/kg^2

T = 2 \pi \sqrt{\frac{(4.73 	imes 10^{20})^3}{6.67 	imes 10^{-11} 	imes 2.5 	imes 10^{41}}}

Calculate numerator:
(4.73 × 10^{20})^3 = 1.06 × 10^{62}

Denominator:
6.67 × 10^{-11} × 2.5 × 10^{41} = 1.6675 × 10^{31}

T = 2 \pi \sqrt{\frac{1.06 	imes 10^{62}}{1.6675 	imes 10^{31}}} = 2 \pi \sqrt{6.36 	imes 10^{30}} = 2 \pi 	imes 7.97 	imes 10^{15} = 5.01 	imes 10^{16} s

Convert seconds to years:
1 year = 3.15 × 10^{7} s

T = \frac{5.01 	imes 10^{16}}{3.15 	imes 10^{7}} = 1.59 	imes 10^{9} years

Therefore, the star takes approximately 1.6 billion years to complete one revolution around the galactic center. — Using Newtonian gravity and orbital mechanics, the orbital period is calculated from the mass enclosed within the star's orbit and its distance from the center, converting units appropriately.

Question 6

(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.

Accepted Answer

(a) The total energy of an orbiting satellite is negative of its kinetic energy. This is because total energy E = K + U, and for a satellite in circular orbit, K = -U/2, so E = -K.

(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 out of Earth's influence. This is because the satellite already has kinetic energy due to its orbital motion. — In orbital mechanics, total energy is negative and equal in magnitude to kinetic energy. An orbiting satellite requires less additional energy to escape Earth's gravity than a stationary object at the same altitude because it already possesses kinetic energy.

Question 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?

Accepted Answer

(a) No, escape speed does not depend on the mass of the body because gravitational potential energy and kinetic energy both scale with mass, which cancels out.

(b) Yes, escape speed depends on the location because gravitational potential varies with distance from Earth's center.

(c) No, escape speed does not depend on the direction of projection; it depends only on the magnitude of velocity needed to escape.

(d) Yes, escape speed depends on the height from which the body is launched; it decreases with increasing height because gravitational potential energy is higher at greater heights. — Escape speed is derived from energy conservation and depends on gravitational potential at the launch point, not on the mass or direction of the body.

Question 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.

Accepted Answer

(a) Linear speed: No, it varies; the comet moves faster near perihelion and slower near aphelion.

(b) Angular speed: No, it varies due to the elliptical orbit.

(c) Angular momentum: Yes, it remains constant because there is no external torque.

(d) Kinetic energy: No, it varies with speed.

(e) Potential energy: No, it varies with distance from the Sun.

(f) Total energy: Yes, it remains constant (neglecting mass loss) because the system is conservative. — In elliptical orbits, speed and potential energy vary, but total mechanical energy and angular momentum are conserved.

Question 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.

Accepted Answer

(a) Swollen feet: No, astronauts experience fluid shift towards the upper body.

(b) Swollen face: Yes, due to fluid redistribution in microgravity.

(c) Headache: Yes, can occur due to fluid shifts and other factors.

(d) Orientational problem: Yes, due to absence of gravity cues affecting vestibular system. — In microgravity, fluids shift towards the head causing facial puffiness, headaches, and disorientation due to vestibular effects.

Question 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 7.11) (i) a, (ii) b, (iii) c, (iv) 0.

Accepted Answer

The gravitational intensity at the centre of a hemispherical shell of uniform mass density is directed towards the flat face of the hemisphere, i.e., along the axis perpendicular to the base. Therefore, the correct answer is (ii) b. — Due to symmetry, the gravitational field contributions from the curved surface cancel in directions perpendicular to the axis, resulting in a net field directed along the axis towards the flat face.

Question 11

For the above problem, the direction of the gravitational intensity at an arbitrary point  $\mathbf{P}$  is indicated by the arrow (i) d, (ii) e, (iii) f, (iv) g.

Accepted Answer

The direction of gravitational intensity at an arbitrary point P near a hemispherical shell depends on the vector sum of contributions from all mass elements. Based on the figure (Fig 7.11), the correct direction is indicated by arrow (iii) f. — The gravitational field points towards the mass distribution; at point P, the net field direction is given by the vector sum of all contributions, which corresponds to arrow f in the figure.

Question 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  $= 2 	imes 10^{30} \, \mathrm{kg}$ , mass of the earth  $= 6 	imes 10^{24} \, \mathrm{kg}$ . Neglect the effect of other planets etc. (orbital radius  $= 1.5 	imes 10^{11} \, \mathrm{m}$ ).

Accepted Answer

Let the distance from the Earth's center where gravitational force is zero be x (measured towards the Sun).

Distance between Earth and Sun = R = 1.5 × 10^{11} m

At point x, gravitational force due to Earth = gravitational force due to Sun

\frac{G M_E}{x^2} = \frac{G M_S}{(R - x)^2}

Cancel G:

\frac{M_E}{x^2} = \frac{M_S}{(R - x)^2}

Take square root:

\frac{\sqrt{M_E}}{x} = \frac{\sqrt{M_S}}{R - x}

Cross multiply:

\sqrt{M_E} (R - x) = \sqrt{M_S} x

\sqrt{M_E} R - \sqrt{M_E} x = \sqrt{M_S} x

Bring terms with x together:

\sqrt{M_E} R = x (\sqrt{M_S} + \sqrt{M_E})

Solve for x:

x = \frac{\sqrt{M_E} R}{\sqrt{M_S} + \sqrt{M_E}}

Calculate:

\sqrt{M_E} = \sqrt{6 	imes 10^{24}} = 2.45 	imes 10^{12}

\sqrt{M_S} = \sqrt{2 	imes 10^{30}} = 1.414 	imes 10^{15}

x = \frac{2.45 	imes 10^{12} 	imes 1.5 	imes 10^{11}}{1.414 	imes 10^{15} + 2.45 	imes 10^{12}} \approx \frac{3.675 	imes 10^{23}}{1.416 	imes 10^{15}} = 2.59 	imes 10^{8} m

Therefore, the gravitational force on the rocket is zero at approximately 2.6 × 10^{8} m from the Earth's center towards the Sun. — Equating gravitational forces from Earth and Sun and solving for the distance where they balance gives the point of zero net gravitational force.

Question 13

How will you 'weigh the sun', that is estimate its mass? The mean orbital radius of the earth around the sun is  $1.5 	imes 10^{8} \, \mathrm{km}$ .

Accepted Answer

We can estimate the mass of the Sun by using the orbital parameters of the Earth.

Given:
Orbital radius, r = 1.5 × 10^{8} km = 1.5 × 10^{11} m
Orbital period, T = 1 year = 3.15 × 10^{7} s

Using Newton's form of Kepler's third law:

M = \frac{4 \pi^2 r^3}{G T^2}

Substitute values:

G = 6.67 × 10^{-11} N·m^2/kg^2

Calculate M:

M = \frac{4 \pi^2 (1.5 	imes 10^{11})^3}{6.67 	imes 10^{-11} (3.15 	imes 10^{7})^2}

Calculate numerator:
(1.5 × 10^{11})^3 = 3.375 × 10^{33}

4 π^2 × 3.375 × 10^{33} ≈ 1.33 × 10^{35}

Calculate denominator:
6.67 × 10^{-11} × 9.92 × 10^{14} = 6.61 × 10^{4}

Therefore,
M = 1.33 × 10^{35} / 6.61 × 10^{4} ≈ 2.0 × 10^{30} kg

Thus, the mass of the Sun is approximately 2 × 10^{30} kg. — Using Earth's orbital radius and period, and Newton's law of gravitation, the Sun's mass is calculated from the centripetal force balance.

Question 14

A saturn year is 29.5 times the earth year. How far is the saturn from the sun if the earth is  $1.50 	imes 10^{8} \, \mathrm{km}$  away from the sun?

Accepted Answer

Given:
Saturn's orbital period, T_S = 29.5 × Earth's orbital period, T_E
Earth's orbital radius, R_E = 1.50 × 10^{8} km

Using Kepler's third law:

\frac{T_S^2}{T_E^2} = \frac{R_S^3}{R_E^3}

=> (29.5)^2 = \frac{R_S^3}{(1.50 	imes 10^{8})^3}

=> R_S^3 = (29.5)^2 	imes (1.50 	imes 10^{8})^3

Calculate:

(29.5)^2 = 869.0

Therefore,

R_S = (869.0)^{1/3} 	imes 1.50 	imes 10^{8} km

Cube root of 869 ≈ 9.54

So,

R_S ≈ 9.54 	imes 1.50 	imes 10^{8} = 1.43 	imes 10^{9} km

Hence, Saturn is approximately 1.43 × 10^{9} km away from the Sun. — Kepler's third law relates the square of the orbital period to the cube of the orbital radius, allowing calculation of Saturn's distance from the Sun.

Question 15

A body weighs  $63\mathrm{N}$  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?

Accepted Answer

Given:
Weight on surface, W = 63 N
Height above surface, h = (1/2) R

Weight at height h, W_h = mg_h

Acceleration due to gravity at height h:

g_h = g \left(\frac{R}{R + h}ight)^2

Since h = R/2,

g_h = g \left(\frac{R}{R + R/2}ight)^2 = g \left(\frac{R}{1.5 R}ight)^2 = g \left(\frac{2}{3}ight)^2 = g 	imes \frac{4}{9}

Weight at height h:

W_h = W 	imes \frac{4}{9} = 63 	imes \frac{4}{9} = 28 N

Therefore, the gravitational force on the body at height h is 28 N. — Weight decreases with height according to the inverse square law of distance from Earth's center.

Gravitation

Gravitation — Study Notes

7.1 Introduction

7.2 Kepler's laws

7.3 Universal law of gravitation

Practice Questions — Gravitation

All 7 Chapters in Physics Part-I