Concepts:
Acceleration Due to Gravity: The acceleration due to Earth’s gravity is given by g = GM/R².
Gravitational Potential Energy: The potential energy of an object in a gravitational field is given by U = -GMm/r.
Key Points:
* Gravitational force is a conservative force.
* Gravitational potential energy is always negative.
* Kepler’s laws were empirical observations.
* Newton’s law of universal gravitation provides a theoretical explanation for Kepler’s laws.
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, Gravitational forces are independent of the medium and cannot be shielded.
(b) Yes, if the spaceship is sufficiently large, its own gravitational field might become detectable, leading to measurable variations in the gravitational field within the spaceship.
(c) Tidal forces decrease with the cube of the distance, while gravitational force decreases with the square of the distance. Due to its closer proximity to Earth, the Moon exerts a stronger tidal force on Earth’s oceans, despite the Sun’s stronger overall gravitational pull.
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 with increasing altitude.
(b) Decreases with increasing depth (assuming uniform Earth density).
(c) Independent of the mass of the body.
(d) More accurate than the formula mg(r2 – r1).
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, T_e = 1 year; T_p = T_e/2 = 1/2 year; r_e = 1 AU.
Applying Kepler’s Third Law, we get the relation:
(T_p/T_e)^2 = (r_p/r_e)^3
Substituting the given values:
(1/2)^2 = (r_p/1)^3
Solving for r_p:
r_p = (1/2)^(2/3)
≈ 0.63 AU
Therefore, the orbital size of the planet is approximately 0.63 AU.
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, we have:
* Orbital period, T₁ = 1.769 days = 1.769 × 24 × 60 × 60 seconds
* Orbital radius, r₁ = 4.22 × 10⁸ m
Using Kepler’s Third Law, the mass of Jupiter (M₁) can be calculated as:
M₁ = (4π²r₁³) / (GT₁²)
Similarly, for the Earth-Sun system:
* Orbital period, T₂ = 1 year
= 365.25 × 24 × 60 × 60 seconds
* Orbital radius, r₂ = 1 AU
≈ 1.496 × 10¹¹ m
The mass of the Sun (M₂) can be calculated using the same formula:
M₂ = (4π²r₂³) / (GT₂²)
To find the ratio of the masses of Jupiter and the Sun, we can divide the two equations:
M₁ / M₂ = (r₁³T₂²) / (r₂³T₁²)
Substituting the values:
M₁ / M₂ ≈ 1/1046
Therefore, the mass of Jupiter is approximately 1/1000th the mass of the Sun.
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 = 50000 ly = 50000 × 9.46 × 10¹⁵ m = 4.73 × 10²⁰ m
M = 2.5 × 10¹¹ solar masses = 2.5 × 10¹¹ × (2 × 10³⁰ kg) = 5.0 × 10⁴¹ kg
We know that the orbital period T of a celestial body orbiting a massive object of mass M at a distance r is given by Kepler’s Third Law:
T² = (4π²r³) / (GM)
Substituting the given values:
T² = (4π² × (4.73 × 10²⁰)³) / (6.67 × 10⁻¹¹ × 5.0 × 10⁴¹)
Solving for T:
T ≈ 1.12 × 10¹⁶ seconds
we divide by the number of seconds in a year, For convert this to years,
T ≈ 1.12 × 10¹⁶ seconds / (365.25 days/year × 24 hours/day × 3600 seconds/hour)
≈ 3.55 × 10⁸ years
Therefore, the star takes approximately 355 million years to complete one revolution around the galactic center.
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) kinetic energy ; If the zero point of potential energy is at infinity, the total mechanical energy of an orbiting satellite is equal to its kinetic energy. Since the total mechanical energy of a bound system (like a satellite orbiting the Earth) is negative, the kinetic energy of the satellite must also be negative.
(b) less ; An orbiting satellite already possesses kinetic energy due to its orbital motion. This kinetic energy can be used to help overcome Earth’s gravitational pull. In contrast, a stationary object at the same height has zero kinetic energy and thus requires additional energy to escape Earth’s gravity.
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 of a body from the Earth, v_e, is given by the formula:
v_e = √(2GM/R) = √(2gR)
Analyzing this equation, we can conclude that:
(a) The escape speed does not depend on the mass of the body being launched.
(b) The escape speed depends on the mass (M) and radius (R) of the Earth, which are constant. Therefore, it does not depend on the location from where the body is projected.
(c) The escape speed is a scalar quantity and does not depend on the direction of projection.
(d) The escape speed depends on the gravitational potential at the point of projection. As the gravitational potential decreases with increasing height, the escape speed also decreases with increasing height.
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) Linear speed:
No. As the comet moves closer to the Sun, its speed increases due to conservation of angular momentum. When it’s farther away, its speed decreases.
b) Angular speed:
No. Similar to linear speed, angular speed also varies. When the comet is closer to the Sun, it covers a larger angle in less time, resulting in a higher angular speed.
c) Angular momentum:
Yes. Angular momentum is conserved in the absence of external torques. Since the comet’s orbit is mainly influenced by the Sun’s gravitational force, which acts along the radial direction, there’s no external torque. Thus, angular momentum remains constant.
d) Kinetic energy:
o. Kinetic energy depends on the square of the speed. As the speed varies, so does the kinetic energy.
e) Potential energy:
No. Gravitational potential energy depends on the distance between the comet and the Sun. As the distance changes, so does the potential energy.
f) Total energy:
Yes. In a closed orbit influenced only by conservative forces (like gravity), the total mechanical energy (kinetic + potential) remains constant.
Therefore, the correct options are:
(c) angular momentum
(f) total energy
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 : (b) swollen face is the most likely symptom to afflict an astronaut in space.
Here’s why:
(a) Swollen feet: In zero gravity, the blood doesn’t pool in the feet, so this is unlikely.
(c) Headache: While headaches can occur in space due to various reasons, they are not directly related to the effects of weightlessness.
(d) Orientation problems: Astronauts do experience spatial disorientation initially, but this is more related to the brain adapting to the new environment rather than a physical symptom.
The main reason for a swollen face is the redistribution of fluids in the body due to weightlessness. Blood and other fluids tend to pool in the upper body, leading to facial swelling.
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.
Ans : The gravitational potential inside a hollow spherical shell is constant. Therefore, the gravitational field (or intensity) at any point inside the shell is zero. This means that the net gravitational force acting on a particle placed inside the shell is zero.
If we remove the upper hemisphere, the symmetry is broken. The net gravitational force on a particle at point P will now be directed downwards towards the remaining lower hemisphere. Consequently, the gravitational intensity at point P will also point downwards, which corresponds to option (iii).
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 : As explained in the previous problem, the gravitational field inside a hollow spherical shell is zero. However, if we remove a portion of the shell, the symmetry is broken. The net gravitational force on a point inside the remaining shell will now be directed towards the denser region. In this case, the gravitational field at point P will be directed along e. Hence option (ii) correct.
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 : Let’s denote the distance between the rocket and the Earth as ‘x’.
The gravitational force on the rocket due to the Earth is:
F_earth = G * (M_earth * m) / x²
The gravitational force on the rocket due to the Sun is:
F_sun = G * (M_sun * m) / (r – x)²
where:
* M_E = mass of the Earth.
* M_S = mass of the Sun
* m = mass of the rocket
* r= Earth-Sun distance
For the net gravitational force to be zero, the two forces must be equal:
F_earth = F_sun
G * (M_earth * m) / x² = G * (M_sun * m) / (r – x)²
Simplifying and substituting the values:
(6 × 10²⁴) / x² = (2 × 10³⁰) / (1.5 × 10¹¹ – x)²
Solving this equation for x, we find that the distance from the Earth’s center where the net gravitational force is zero is approximately 2.6 × 10⁸ meters.
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 : Let M be the mass of the Sun, and m be the mass of the Earth.
**Using Newton’s Law of Gravitation:**
The gravitational force between the Sun and the Earth is given by:
F = GMm/R²
**Using the concept of centripetal force:**
The Earth’s orbital motion can be considered as circular motion. The centripetal force required for this motion is provided by the gravitational force:
F = mv²/R = mRω²
Equating the two forces:
GMm/R² = mRω²
Simplifying and substituting ω = 2π/T (where T is the Earth’s orbital period), we get:
M = 4π²R³/GT²
Now, we can substitute the given values:
* R = 1.5 × 10¹¹ m
* T = 365.25 days
= 365.25 × 24 × 60 × 60 seconds
* G = 6.67 × 10⁻¹¹ Nm²/kg²
Calculating M, we find that the mass of the Sun is approximately 2.0 × 10³⁰ kg.
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 : Kepler’s Third Law : Mathematically, this can be expressed as:
T² ∝ a³
For Saturn and Earth, we can write:
(T_Saturn / T_Earth)³ = (a_Saturn / a_Earth)³
Given:
* T_Saturn = 29.5 years
* T_Earth = 1 year
* a_Earth = 1.5 × 10⁸ km
Substituting these values and solving for a_Saturn, we find:
a_Saturn ≈ 1.43 × 10⁹ km
Therefore, Saturn is approximately 1.43 billion kilometers away from the Sun.
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’s denote:
m = The mass of the body
Let R = radius of the Earth.
Let g denote The acceleration due to gravity at the Earth’s surface
We know that the weight of the body on the Earth’s surface is given by:
W = mg = 63 N
Now, the gravitational force on the body at a height R/2 above the Earth’s surface is:
F = G * (M_earth * m) / (R + R/2)²
where:
* G = universal gravitational constant
* M_earth = mass of the Earth
We can relate the weight W and the acceleration due to gravity g:
W = G * (M_earth * m) / R²
Dividing the two equations, we get:
F/W = (R/1.5R)² = 4/9
Therefore, the gravitational force on the body at the given height is:
F = (4/9) * W = (4/9) * 63 N
= 28 N
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 : Let’s denote:
m = The mass of the body
Let R = radius of the Earth.
Let g denote The acceleration due to gravity at the Earth’s surface
We know that the weight of the body on the Earth’s surface is given by:
W = mg = 250 N
Now, the acceleration due to gravity at a depth d below the Earth’s surface is given by:
g’ = g(1 – d/R)
At a depth of R/2, g’ becomes:
g’ = g(1 – (R/2)/R) = g/2
Therefore, the weight of the body at a depth of R/2:
W’ = mg’ = m(g/2) = W/2 = 250 N / 2
= 125 N
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 : we can apply the law of conservation of mechanical energy, for solve this problem
*Initial Energy:
Kinetic energy: (1/2)mv²
* Potential energy: -GMm/R
**Final Energy:**
At the maximum height, the rocket momentarily comes to rest, so its kinetic energy becomes 0. Its potential energy will be: -GMm/r, where r is the maximum distance from the Earth’s center.
**Equating initial and final energy:**
(1/2)mv² – GMm/R = -GMm/r
Simplifying and solving for r:
r = GM/(GM/R – (1/2)v²)
Substituting the given values:
r = (6.67 × 10⁻¹¹ Nm²/kg² * 6.0 × 10²⁴ kg) / [(6.67 × 10⁻¹¹ Nm²/kg² * 6.0 × 10²⁴ kg / 6.4 × 10⁶ m) – (1/2) * (5000 m/s)²]
Calculating r, we find that the rocket goes approximately 3.6 × 10⁷ m away from the Earth’s center before returning.
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 v_e be the escape speed from the surface of the Earth, which is 11.2 km/s.
By definition, the escape speed is the minimum speed required for an object to escape the gravitational pull of a celestial body. At this speed, the object’s kinetic energy is equal to its gravitational potential energy:
(1/2)mv_e² = GMm/R
where:
* m = mass of the object
* M = mass of the Earth
* R = radius of the Earth
* G = gravitational constant
Now, if a body is projected with a speed of 3v_e, its final speed (v_f) far away from the Earth can be found using the conservation of energy principle:
(1/2)m(3v_e)² – GMm/R = (1/2)mv_f²
Simplifying and solving for v_f, we get:
v_f = √(9v_e² – v_e²) = √8v_e² = √8 × 11.2 km/s
≈ 31.7 km/s
hence, the final speed of the body far away from the Earth is approximately 31.7 km/s.
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 : Understanding the Problem:**
We are asked to find the energy required to move a satellite from its initial orbit at a height of 400 km to infinity, where the Earth’s gravitational influence becomes negligible.
**Solution:**
**Initial Energy of the Satellite:**
At a height of 400 km, the satellite possesses both potential and kinetic energy.
* **Potential Energy (PE):**
PE = -GMm / (R+h)
where:
* G = gravitational constant
* M = mass of the Earth
* m = mass of the satellite
* R = radius of the Earth
*h stands for altitude of the satellite above the Earth’s surface.
Kinetic Energy (KE):**
For a stable orbit, the kinetic energy is equal to half the magnitude of the potential energy.
KE = GMm / (2(R+h))
Therefore, the total initial energy (E₁) of the satellite is:
E₁ = PE + KE = -GMm / (2(R+h))
Final Energy of the Satellite:
At infinity, the potential energy is zero, and since the satellite will eventually come to rest, its kinetic energy will also be zero. So, the final energy (E₂) is zero.
**Energy Required:**
The energy required to move the satellite out of Earth’s gravitational influence is the difference between the final and initial energies:
Energy required = E₂ – E₁ = 0 – (-GMm / (2(R+h))) = GMm / (2(R+h))
Substituting the given values:
Energy required = (6.67 × 10⁻¹¹ Nm²/kg²) × (6.0 × 10²⁴ kg) × (200 kg) / (2(6.4 × 10⁶ m + 4 × 10⁵ m))
= 5.9*10^9J
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 10^9 km, their speeds are negligible. What is the speed with which they collide ? The radius of each star is 10^4 km. Assume the stars to remain undistorted until they collide.
(Use the known value of G).
Ans : Let M be the mass of each star (2 x 10^30 kg) and r be their initial separation (10^9 km = 10^12 m).
Initially, the system’s potential energy is:
PE_initial = -GM²/r
As the stars approach each other, their potential energy decreases, converting into kinetic energy. At the point of collision, when they are separated by a distance of 2R (the sum of their radii), their potential energy is:
PE_final = -GM²/(2R)
The change in potential energy, ΔPE = PE_final – PE_initial, is equal to the total kinetic energy gained by the stars:
ΔPE = 2 × (1/2)Mv² = Mv²
Therefore, we can equate the change in potential energy to the final kinetic energy and solve for v:
Mv² = GM²(1/r – 1/(2R))
v = √[GM(1/r – 1/(2R))]
Substituting the given values for G, M, r, and R, we can calculate the final speed of the stars at the moment of collision.
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 : The gravitational force between the two spheres, each of mass 100 kg and separated by 1 meter, can be calculated using Newton’s law of gravitation:
F = G(m₁m₂) / r²
≈ 6.674 × 10⁻⁷ N
The gravitational potential at the midpoint between the spheres is the sum of the potentials due to each sphere:
V = -GM₁/r – GM₂/r = -2GM/r
≈ -2.67 × 10⁻⁸ J/kg
An object placed at the midpoint will be in equilibrium as the gravitational forces from both spheres will balance each other. However, this equilibrium is unstable. Any slight displacement will result in a net force pulling the object further away from the equilibrium point.
