Question:

The escape speed of a projectile on the earth’s surface is 11.2 kms^−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.

Answer:

Answer:

1. The escape speed of a projectile on the earth’s surface is 11.2 kms^−1.
2. A body is projected out with thrice this speed, so the speed of the body is 3 x 11.2 kms^−1 = 33.6 kms^−1.
3. Far away from the earth, the speed of the body remains the same, i.e. 33.6 kms^−1.

Question:

Let us assume that our galaxy consists of 2.5×10^11 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.

Answer:

Step 1: Calculate the circumference of the Milky Way. Circumference = π × Diameter = 3.14 × 10^5 ly

Step 2: Calculate the total mass of the Milky Way. Total mass = 2.5 × 10^11 × 1 solar mass = 2.5 × 10^12 solar masses

Step 3: Calculate the centripetal force acting on the star. Centripetal force = (2.5 × 10^12 solar masses) / (50,000 ly)^2

Step 4: Calculate the period of revolution of the star. Period of revolution = (2π × 50,000 ly) / (Centripetal force)

Step 5: Calculate the time taken by the star to complete one revolution. Time taken to complete one revolution = (Period of revolution) / (2π)

Therefore, the star at a distance of 50,000 ly from the galactic centre will take 4.17 × 10^7 years to complete one revolution.

Question:

One of the satellites of Jupiter has an orbital period of 1.769 days and the radius of the orbit is 4.22×10^8m. Show that the mass of Jupiter is about one-thousandth that of the sun.

Answer:

Step 1: Calculate the gravitational force of Jupiter using the equation F=GmM/r^2, where G is the gravitational constant (6.67x10^-11), m is the mass of Jupiter, M is the mass of the sun, and r is the radius of the satellite’s orbit.

Step 2: Calculate the centripetal force of the satellite using the equation F=mv^2/r, where m is the mass of the satellite, v is the velocity of the satellite, and r is the radius of the satellite’s orbit.

Step 3: Set the gravitational force and the centripetal force equal to each other, and solve for the mass of Jupiter: F=GmM/r^2=mv^2/r.

Step 4: Plug in the given values for G, M, r, and v (the velocity can be calculated using the equation v=2πr/T, where T is the orbital period of the satellite).

Step 5: Solve for m, the mass of Jupiter, and compare it to the mass of the sun. The mass of Jupiter is about one-thousandth that of the sun.

Question:

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?

Answer:

a) Yes, the escape speed of a body from the earth depends on the mass of the body. The greater the mass of the body, the greater the escape speed.

b) Yes, the escape speed of a body from the earth depends on the location from where it is projected. The further away from the earth the body is projected, the greater the escape speed.

c) Yes, the escape speed of a body from the earth depends on the direction of projection. The direction of projection affects the speed and angle at which the body is launched, and thus affects the escape speed.

d) No, the escape speed of a body from the earth does not depend on the height of the location from where the body is launched. The height of the location has no effect on the escape speed.

Question:

A rocket is fired ‘vertically’ from the surface of mars with a speed of 2 kms^−1. If 20% of its initial energy is lost due to martian atmospheric resistance, how far (in km) will the rocket go from the surface of mars before returning to it ? Mass of mars = 6.4×10^23 kg; radius of mars = 3395 km; G=6.67×10^−11 Nm^2kg^−2.

Answer:

Step 1: Calculate the gravitational potential energy of the rocket at the surface of Mars:

Gravitational potential energy = (mass of Mars x gravitational constant) / radius of Mars

Gravitational potential energy = (6.4 x 10^23 kg x 6.67 x 10^-11 Nm^2kg^-2) / 3395 km

Gravitational potential energy = 8.08 x 10^10 J

Step 2: Calculate the kinetic energy of the rocket at the surface of Mars:

Kinetic energy = (mass of rocket x speed of rocket^2) / 2

Kinetic energy = (1 kg x (2 kms^-1)^2) / 2

Kinetic energy = 2 x 10^3 J

Step 3: Calculate the total energy of the rocket at the surface of Mars:

Total energy = Gravitational potential energy + Kinetic energy

Total energy = 8.08 x 10^10 J + 2 x 10^3 J

Total energy = 8.08 x 10^10 J

Step 4: Calculate the energy lost due to martian atmospheric resistance:

Energy lost = 20% of initial energy

Energy lost = 20% of 8.08 x 10^10 J

Energy lost = 1.616 x 10^10 J

Step 5: Calculate the final energy of the rocket:

Final energy = Total energy - Energy lost

Final energy = 8.08 x 10^10 J - 1.616 x 10^10 J

Final energy = 6.464 x 10^10 J

Step 6: Calculate the final speed of the rocket:

Final speed = (2 x Final energy)^1/2

Final speed = (2 x 6.464 x 10^10 J)^1/2

Final speed = 1.96 kms^-1

Step 7: Calculate the distance traveled by the rocket before returning to the surface of Mars:

Distance traveled = Final speed x time

Time = (2 x distance traveled) / Final speed

Time = (2 x Distance traveled) / 1.96 kms^-1

Distance traveled = (1.96 kms^-1 x Time) / 2

Time = (2 x Distance traveled) / 1.96 kms^-1

Distance traveled = (1.96 kms^-1 x Time) / 2

Time = (2 x Distance traveled) / 1.96 kms^-1

Distance traveled = (1.96 kms^-1 x Time) / 2

Time = (2 x 3395 km) / 1.96 kms^-1

Time = 3447.959 s

Distance traveled = (1.96 kms^-1 x 3447.959 s) / 2

Distance traveled = 3315.5 km

Question:

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 ?

Answer:

1. The gravitational force between the two spheres is given by the equation F = G * (m1 * m2) / (r^2), where G is the gravitational constant, m1 and m2 are the masses of the two spheres, and r is the distance between them. In this case, the gravitational force is equal to F = G * (100 kg * 100 kg) / (1 m^2) = 6.67 * 10^-11 N.

2. The gravitational potential at the midpoint of the line joining the centres of the spheres is given by the equation V = G * (m1 * m2) / (2 * r), where G is the gravitational constant, m1 and m2 are the masses of the two spheres, and r is the distance between them. In this case, the gravitational potential is equal to V = G * (100 kg * 100 kg) / (2 * 1 m) = 3.33 * 10^-11 J.

3. Yes, an object placed at the midpoint of the line joining the centres of the spheres would be in equilibrium, since the gravitational force and potential are equal in magnitude and opposite in direction.

4. The equilibrium would be stable, since any small perturbation from the equilibrium position would result in a restoring force that would bring the object back to the equilibrium position.

Question:

Which of the following symptoms is not likely to afflict an astronaut in space (a) swollen feet, (b) swollen face, (c) headache, (d) orientational problem.

Answer:

Step 1: Read the question carefully.

Step 2: Analyze the possible answers.

Step 3: Consider which answer does not fit the context of the question.

Step 4: Select the answer that does not fit the context of the question.

Answer: (b) Swollen face

Question:

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 −GMm(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 center of the earth.

Answer:

(a) Decreases (b) Increases (c) Mass of the earth (d) More

Question:

Assuming the earth to be a sphere of uniform mass density, how much would a body weigh (in N) half way down to the center of the earth if it weighed 250 N on the surface ?

Answer:

Step 1: Calculate the radius of the earth. The radius of the earth is 6,371 km.

Step 2: Calculate the mass of the earth. The mass of the earth is 5.98 x 10^24 kg.

Step 3: Calculate the gravitational acceleration at the surface of the earth. The gravitational acceleration at the surface of the earth is 9.8 m/s2.

Step 4: Calculate the gravitational force at the surface of the earth. The gravitational force at the surface of the earth is 2.48 x 10^22 N.

Step 5: Calculate the weight of the body on the surface of the earth. The weight of the body on the surface of the earth is 250 N.

Step 6: Calculate the distance from the surface of the earth to the center of the earth. The distance from the surface of the earth to the center of the earth is 3,185.5 km.

Step 7: Calculate the weight of the body half way down to the center of the earth. The weight of the body half way down to the center of the earth is 125 N.

Question:

How will you ‘weigh the sun’, that is estimate its mass? The mean orbital radius of the earth around the sun is 1.5×10^8 km.

Answer:

1. Estimate the volume of the sun using the mean orbital radius of the earth around the sun. This can be done by using the formula for the volume of a sphere, V = (4/3)πr^3, where r is the radius of the sphere.

2. Calculate the density of the sun using the mass and volume of the sun. This can be done by using the formula for density, D = m/V, where m is the mass and V is the volume.

3. Calculate the mass of the sun using the density and volume of the sun. This can be done by using the formula for mass, m = D*V, where D is the density and V is the volume.

Question:

Assuming the earth to be a sphere of uniform mass density, how much would a body weigh (in N) half way down to the center of the earth if it weighed 250 N on the surface ?

Answer:

Answer: Step 1: Calculate the radius of the Earth. Radius of Earth = 6371 km

Step 2: Calculate the distance from the surface to the center of the Earth. Distance from surface to center of Earth = 6371 km / 2 = 3185.5 km

Step 3: Calculate the gravitational acceleration at the halfway point. Gravitational acceleration at halfway point = 9.8 m/s2

Step 4: Calculate the weight of the body at the halfway point. Weight at halfway point = 250 N x (9.8 m/s2 / 3185.5 km) = 0.0078 N

Question:

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 = 6.0×10^24 kg, radius = 6400 km.

Answer:

Step 1: Calculate the acceleration due to gravity at the site of the satellite.

Gravitational acceleration = (G*M)/(R + h)^2

G = Universal Gravitational Constant (6.6710^-11 Nm^2/kg^2) M = Mass of the Earth (6.010^24 kg) R = Radius of the Earth (6400 km) h = Height of the satellite (36,000 km)

Gravitational acceleration = (6.6710^-116.0*10^24)/(6400 + 36,000)^2

Gravitational acceleration = 0.0008 m/s^2

Step 2: Calculate the potential energy at the site of the satellite.

Potential Energy = (GMm)/R

G = Universal Gravitational Constant (6.6710^-11 Nm^2/kg^2) M = Mass of the Earth (6.010^24 kg) m = Mass of the satellite (1 kg) R = Radius of the Earth (6400 km)

Potential Energy = (6.6710^-116.010^241)/6400

Potential Energy = -9.48*10^14 Joules

Question:

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 = 2×10^30 kg; mass of mars = 6.4×10^23 kg; Radius of mars = 3395 km; Radius of the orbit of mars = 2.28×108 km G=6.67×10^−11 Nm^2kg^−2.

Answer:

1. Calculate the gravitational force between the spaceship and the sun: F = G*(mass of the sun*mass of the spaceship)/(radius of the orbit of mars)^2 F = 6.67×10^−11 Nm^2kg^−2 * (2×10^30 kg * 1000 kg)/(2.28×108 km)^2 F = 2.92×10^13 N

2. Calculate the gravitational force between the spaceship and Mars: F = G*(mass of Mars*mass of the spaceship)/(radius of Mars)^2 F = 6.67×10^−11 Nm^2kg^−2 * (6.4×10^23 kg * 1000 kg)/(3395 km)^2 F = 1.93×10^7 N

3. Calculate the energy required to launch the spaceship out of the solar system: E = F * (radius of the orbit of Mars) E = (2.92×10^13 N + 1.93×10^7 N) * (2.28×108 km) E = 1.37×10^22 J

Question:

A rocket is fired vertically with a speed of 5 kms^−1 from the Earth’s surface. How far from the Earth does the rocket go before returning to the Earth ? Mass of the earth =6.0×10^24 kg; mean radius of the radius of the Earth =6.4×10^6 m; G=6.67×10^−11 Nm^2kg^−2.

Answer:

1. Calculate the gravitational force of attraction between the rocket and the Earth: F = G * (m1 * m2) / r^2 F = 6.67 * 10^-11 * (6.0 * 10^24 * 1) / (6.4 * 10^6)^2 F = 0.0098 N

2. Calculate the acceleration due to gravity: a = F/m a = 0.0098 / 1 a = 0.0098 m/s^2

3. Calculate the time it takes for the rocket to reach its maximum height: t = v/a t = 5 / 0.0098 t = 507.14 s

4. Calculate the maximum height reached by the rocket: h = vt - (1/2)at^2 h = 5 * 507.14 - (1/2) * 0.0098 * 507.14^2 h = 2.5355 * 10^6 m

5. Calculate the distance from the Earth when the rocket returns to the Earth: d = 2h d = 2 * 2.5355 * 10^6 d = 5.071 * 10^6 m

Question:

A rocket is fired from the earth towards the sun. At what distance from the earths centre is the gravitational force on the rocket zero? Mass of the sun = 2×10^30kg, mass of the earth = 6×10^24kg. Neglect the effect of other planets etc. (orbital radius = 1.5×10^11m).

Answer:

1. Calculate the gravitational force between the sun and the earth:

F = G*((M_sun*M_earth)/r^2)

F = (6.6710^-11)((210^30610^24)/(1.510^11)^2)

F = 3.8*10^22 N

2. Calculate the distance from the earth’s centre at which the gravitational force on the rocket is zero:

F = G*((M_sun*M_rocket)/r^2)

0 = (6.6710^-11)((210^30m_rocket)/r^2)

r = (210^30m_rocket)/(6.6710^-110)

r = ∞ (infinity)

Therefore, the distance from the earth’s centre at which the gravitational force on the rocket is zero is infinity.

Question:

A saturn year is 29.5 times the earth year. How far is the saturn from the sun if the earth is 1.50×10^8km away from the sun ?

Answer:

Step 1: Calculate the number of Earth years in one Saturn year.

1 Saturn year = 29.5 Earth years

Step 2: Calculate the distance of Saturn from the Sun in terms of Earth years.

Distance of Saturn from the Sun = 1.50×10^8km x 29.5

Step 3: Calculate the distance of Saturn from the Sun in kilometers.

Distance of Saturn from the Sun = 4.425×10^9km

Question:

A body weighs 63N on the surface of the earth. What is the gravitational force (in N) on it due to the earth at a height equal to half the radius of the earth ?

Answer:

Step 1: Calculate the radius of the Earth. Radius of the Earth = 6,371 km

Step 2: Calculate the height of the body. Height = Half the radius of the Earth = 3,185.5 km

Step 3: Calculate the gravitational force on the body. Gravitational Force (F) = (Gravitational Constant x Mass of the Earth x Mass of the Body) / (Radius of the Earth)2 F = (6.67 x 10-11 x 5.97 x 1024 x 63) / (6,371)2 F = 9.81 N

Question:

Two stars each of one solar mass (=2×10^30 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).

Answer:

Step 1: Calculate the force of gravity between the two stars. F = Gm1m2/r2

Step 2: Calculate the acceleration of the stars due to gravity. a = Gm1m2/r2

Step 3: Calculate the speed of the stars when they are 10^9 km apart. v = sqrt(2Gm1*m2/r)

Step 4: Calculate the speed of the stars when they collide. v = sqrt(2Gm1*m2/r1)

where G = 6.67 x 10-11 Nm2/kg2 m1 = m2 = 2 x 1030 kg r = 10^9 km r1 = 10^4 km

Question:

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 ?

Answer:

Step 1: Understand the question. The question is asking what the orbital size of a planet would be if it went around the sun twice as fast as the Earth.

Step 2: Calculate the orbital size. The orbital size of a planet is determined by its orbital period, which is the time it takes to complete one orbit around the sun. Since the hypothetical planet is going around the sun twice as fast as the Earth, its orbital period would be half that of the Earth, which is 365.25 days. Therefore, the orbital size of the hypothetical planet would be half that of the Earth.

Question:

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 earths gravitational influence is more/less than the energy required to project a stationary object at the same height (as the satellite) out of earths influence.

Answer:

Answer: (b) The energy required to launch an orbiting satellite out of earths gravitational influence is more than the energy required to project a stationary object at the same height (as the satellite) out of earths influence.

Question:

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.

Answer:

(a) No, the comet does not have a constant linear speed throughout its orbit. Its speed changes as it moves closer and farther away from the Sun.

(b) No, the comet does not have a constant angular speed throughout its orbit. Its angular speed changes as it moves closer and farther away from the Sun.

(c) Yes, the comet has a constant angular momentum throughout its orbit.

(d) No, the comet does not have a constant kinetic energy throughout its orbit. Its kinetic energy changes as it moves closer and farther away from the Sun.

(e) No, the comet does not have a constant potential energy throughout its orbit. Its potential energy changes as it moves closer and farther away from the Sun.

(f) No, the comet does not have a constant total energy throughout its orbit. Its total energy changes as it moves closer and farther away from the Sun.

Question:

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 =6.0×10^24 kg; radius of the earth =6.4×10^6m; G = 6.67×10^−11 Nm^2kg^−2.

Answer:

1. Calculate the gravitational force between the satellite and the earth: F = (G * 6.0x10^24 kg * 200 kg) / (6.4x10^6 m)^2 F = 1.83x10^17 N

2. Calculate the potential energy of the satellite at its current height: PE = (F * 400 km) PE = 7.32x10^19 J

3. Calculate the kinetic energy required to escape the earth’s gravitational influence: KE = PE KE = 7.32x10^19 J

Question:

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 = 2×10^30kg).

Answer:

Step 1: Calculate the mass of the star. Mass of the star = 2.5 x 2 x 10^30 kg = 5 x 10^30 kg

Step 2: Calculate the radius of the star. Radius of the star = 12 km = 12 x 10^3 m

Step 3: Calculate the gravitational acceleration on the surface of the star. Gravitational acceleration = (G x Mass of the star)/(Radius of the star)^2 Gravitational acceleration = (6.67 x 10^-11 x 5 x 10^30)/(12 x 10^3)^2 Gravitational acceleration = 2.22 x 10^4 m/s^2

Step 4: Calculate the centripetal acceleration required for an object to remain stuck to its surface. Centripetal acceleration = (Velocity of the star)^2/Radius of the star Centripetal acceleration = (1.2 x 2 x pi)^2/(12 x 10^3) Centripetal acceleration = 9.55 x 10^3 m/s^2

Step 5: Compare the gravitational acceleration and the centripetal acceleration. Gravitational acceleration > Centripetal acceleration

Conclusion: An object placed on the equator of the star will remain stuck to its surface due to gravity.