## NCERT Solutions for Class 11 Physics Chapter 8 - GravitationStudents should use the NCERT Answers for Class 11 Physics Chapter 8 Gravitation as important resources if they want to do well in the Class 11 exam. These solutions are comparable to the professionals consulting several texts. The CBSE Syllabus 2022-23 is strictly followed in these solutions. They provide students with example papers, relevant questions from past years' question papers, and answers to the questions found in the textbooks. The experts at Javatpoint created the NCERT Answers for Class 11 Physics Chapter 8 to assist students in structuring their responses to challenging exam questions. In our lives, we know that all material objects are attracted towards the earth. Anything which we throw up falls down. Climbing a hill is more tiring than going downhill. Want to know more about Gravitation? Not able to answer the questions from NCERT Textbook? Is understanding the concepts difficult? Javatpoint is the one-stop solution for all students' needs. ## NCERT Solutions for Class 11 Physics Chapter 8
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?
a.) The gravitational force is unaffected by the make up of the material and its medium. The two parameters that directly affect gravitational force are the body's mass and the separation between two bodies. A body cannot thus resist the effects of gravity. b.) The size of the spacecraft plays a crucial role in how the astronauts experience gravity. The gravitational effect of a spacecraft's increasing size can be measured. Therefore, the astronaut wants to find gravity. c.) The distance plays a significant role in the tidal effect. The tidal effect is impacted by gravitational force in a way that is inverse to the square of the distance. The tidal effect changes with the square root of the inverse of the distance. More than twice as far as the moon is from the ocean as the sun is from the earth. As a result, the moon's pull on the tides is stronger than the sun's.
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, is more/less accurate than the formula mg = (r for the difference of potential energy between two points r
a.) decreases The acceleration due to gravity at depth his given as, where, Radius of earth is R Hence, the aforementioned equation demonstrates that acceleration caused by gravity reduces as altitude increases. b.) decreases The acceleration due to gravity at depth his given as follows, where, depth from earth surface is d. So, it is evident from the relationship above that acceleration caused by gravity declines as depth increases. c.) mass of the body Acceleration due to gravity of body of mass m is given as, g = GMR where, universal gravitational constant is G, mass of earth is M and radius of earth is R. Thus the above equation shows that acceleration due to gravity is independent of mass of the body. d.) more The difference in potential energy at radius r Thus, this formula is more accurate than the formula, mg = (r
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?
Lesser by a factor of 0.63 1 Year.T Radius of the earth i.e.its orbital radius = R The amount of time it takes for a planet to make one revolution around the Sun, The planet's orbital radius will therefore be 0.63 times smaller than the Earth's.
One of the satellites of Jupiter, has an orbital period of 1.769 days, and the radius of the orbit is 4.22 × 10
Let us assume that our galaxy consists of 2.5 × 10
Mass of our galaxy Milky Way,M = 2.5 × 10 Solar Mass=Mass of the Sun = 2.0 × 10 Mass of our Galaxy,M = 2.5 × 10 Diameter of the Milky Way,d = 10 Radius of the Mily Way, r = 5 × 1 light year = 9.46 × 10 ∴ r = 5 × 10 = 4.73 × 10 A star's period is determined by the relation since it spins around the Milky Way's galactic centre:
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.
a.) A satellite's kinetic energy is usually positive, although its potential energy could be negative. The total energy of the satellite is equal to the product of its kinetic and potential energies. The total energy of the Earth-satellite system is negative since it is a bound system. As a result, an orbiting satellite's total energy is negative compared to its kinetic energy if the potential energy zero is at infinity. b.) The gravitational potential energy of the satellite is zero at infinity. The total energy of the satellite is negative since it is a bound system with the Earth. An orbiting satellite's total energy is therefore the inverse of its kinetic energy at infinity. A specific amount of energy is acquired by an orbiting satellite, allowing it to spin around the Earth. Its orbit gives forth this energy. Compared to an originally energy-empty stationary object on the surface of the Earth, it takes considerably less energy to move away from the gravitational pull of the planet. As a result, less energy is needed to launch an orbiting satellite out of the gravitational pull of the planet than to raise a stationary object to the same height.
Does the escape speed of a body from the earth depend on - the mass of the body,
- the location from where it is projected,
- the direction of projection,
- the height of the location from where the body is launched?
a.) Considering the relation of escape velocity, Thus, we can say that the escape velocity has no relation with mass of the body. b.) Considering the relation of escape velocity, Although while g fluctuates with latitude, its value on the surface of the earth only varies by a very small amount. As a result, it is possible to say that escape velocity is independent of location. c.) Considering the relation of escape velocity, The escape velocity is not dependent on the direction of the projection d.) Considering the relation of escape velocity, The body's escape velocity is determined by the gravitational potential at the launch location. Given that this potential is somewhat influenced by the latitude and height of the site. As a result, the escape velocity is somewhat dependent upon the height of the area where the body is launched.
A comet orbits the Sun in a highly elliptical orbit. Does the comet have a constant - linear speed
- angular speed
- angular momentum
- kinetic energy
- potential energy
- total energy throughout its orbit?
Neglect any mass loss of the comet when it comes very close to the Sun.
- The linear speed of a comet on an elliptical orbit is not constant because an elliptical orbit causes a continuous change in the comet's distance from the sun.
- The angular speed of the comet varies during the orbit because different angles are traversed in equal intervals of time.
- Due to angular momentum conservation in the absence of torque, the comet's angular momentum remains constant throughout.
- Kinetic energy is not constant since the linear speed varies depending on where in the orbit you are.
- Because the comet is not at the same distance from the sun, the potential energy differs at various locations (the orbit is not circular).
- Throughout its motion, the comet's overall energy doesn't change.
As a result, the comet's linear speed, angular speed, kinetic energy, and potential energy are not consistent throughout its orbit, but its angular momentum and total energy are.
Which of the following symptoms is likely to afflict an astronaut in space - swollen feet,
- swollen face,
- headache,
- orientational problem.
(b), (c), and (d) - Due to gravity's pull, legs hold a body's full bulk when it is standing. Because there is no gravitational pull in space, an astronaut feels as though they are floating. As a result, an astronaut's swollen feet have no effect on him or her in space.
- Generally speaking, the perceived weightlessness in space is what causes a bloated face. A person's face is made up of sense organs like their eyes, ears, nose, and mouth. An astronaut in orbit may have this condition.
- Mental tension might result in headaches. An astronaut's performance in space may be impacted.
- There are various orientations in space. Hence, an astronaut in space may experience orientational issues.
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 ) - a,
- b,
- c,
- 0
In a sphere, the gravitational potential is the same everywhere. The gravitational potential gradient's intensity is therefore zero. As a result, the gravitational force exerted by the Earth's gravity on any particle at any location inside the spherical shell is symmetrical. In the illustration, a spherical shell's top half has been removed. A particle near the centre of the sphere will therefore experience a downward-directed net gravitational pull. Due to the fact that the gravitational force per unit mass determines its intensity, the gravitational intensity will always point downward, or along c. Option (iii) is correct as a result.
For the above problem, the direction of the gravitational intensity at an arbitrary point P is indicated by the arrow - d,
- e,
- f,
- g
In the illustration, a spherical shell's top half has been removed. Hence, a particle will experience a downward-directed net gravitational force at any point P. Because to the fact that the gravitational force per unit mass determines its intensity, the gravitational intensity will always point downward, or along e. Option (ii) is correct as a result.
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 10
Mass of the sun,M Mass of the sun,M Orbital radius,r = 1.5 × 10 Mass of the rocket = m The point at which the gravitational pull on satellite P is zero, let x be the distance from the Earth's centre. The gravitational forces exerted on satellite P by the Sun and the Earth are equivalent to those described by Newton's law of gravitation as follows:
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
Around the sun,the orbital radius of the earth is,r = 1.5 × 10 Amount of time taken to complete 1 revolution by earth around the sun is, 1 Year = 365.25 days = 365.25 × 24 × 60 × 60 s Universal gravitational constant,G = 6.67 × 10 Thus we can calculate the mass of the sun by using the following relation,
A Saturn year is 29.5 times the Earth year. How far is Saturn from the Sun if the Earth is 1.50 x 10
One year of Saturn is equal to 29.5 times Earth Year Let 1 year of time at Earth be = T Thus 1 Year of Saturn,T'=29.5 Thus, following Kepler's law,
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?
Weight of the body,W = 63 N At height h above the Earth's surface,acceleration due to gravity is given by the relation,
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?
A rocket is fired vertically with a speed of 5 km/s 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
Velocity of the rocket,v=5 Km/s=5×10 Mass of the Earth,M Radius of the Earth,R Height reached by the rocket mass,m = h At the surface of the Earth, Total energy of the rocket = Kinetic Energy + Potential Energy
The escape speed of a projectile on the earth's surface is 11.2 km/s. 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.
Escape velocity of the projectile on the Earth's surface,V Speed of projection of body,v = 3Ve = 3 × 11.2 = 33.6 Km/s Let v and v' represent the body's speed at the time of projection and at a distance from the earth, respectively. At the time of projection, Initial kinetic energy of the body = 1/2mv
A satellite orbits the earth at the 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
Height of the satellite,h = 400 Km = 4 × 10 Mass of the Earth,M = 6.0 × 10 Mass of the satellite,m = 200 Kg Radius of the Earth,R The absence of a positive sign denotes the satellite's earthbound position. We refer to this as a satellite's binding energy.
Two stars, each of one solar mass (= 2×10
Mass of each star,M = 2 × 10 Radius of each star,R = 10 Distance between the stars,r = 10 For negligible speed v = 0 Total energy of the two stars separated at a distance r is given by,
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 midpoint 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?
There is a gravitational pull at the mid-point. Two spheres will exert equal and opposing gravitational forces at the midpoint. Hence, there will be no net gravitational force at the midpoint of the line connecting the centres of the spheres. At the midpoint,gravitational potential energy is calculated in the following way, Radius of the sphere,R = 0.10 m Distance between two spheres,r = 1.0 m Mass of each sphere,M = 100 Kg G = 6.67 × 10 The object's equilibrium is present at the mid-point. As the object at the midpoint will experience no net gravitational pull, the system is in equilibrium. The system is in an unstable equilibrium since there isn't a restoring force, and an object that has been moved won't go back to its original position.
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 = 6.0×10
Radius of the Earth = 6400 = 0.64 × 10 Mass of the Earth = 6 × 10 The geostationary satellite is at height h from earth's surface,where h = 3.6 × 10 Thus,the gravitational potential energy possed by geostationary satellite because of the earth's gravity is given by, Thus the gravitational potential energy possessed by the geostationary satellite because of earth's gravity is -9.439 × 10
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
Any mass or item will stay adhered to the surface if the gravitational pull from within is greater than the centrifugal force from without.
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 = 2×10
The mass of the spaceship,m The mass of the Sun,M = 2 × 10 The mass of Mars,m The radius of the orbit of Mars,R = 2.28 × 10 The radius of the Mars,r = 3395 × 10 Universal gravitational constant, G = 6.67 × 10 The spacecraft's potential energy as a result of the Sun's gravitational pull is given by, The potential energy of the spacecraft brought on by Mars's gravitational pull is given by, Total energy of the spaceship, The negative sign indicates that satellite is bound to the system. The amount of power needed to send the spacecraft beyond the solar system is equal to (total energy of the spaceship).
A rocket is fired 'vertically' from the surface of mars with a speed of 2 km/s. 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 = 6.4×10
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 × 10 The radius of Mars,R = 3395 = 3.395 × 10 20% of the kinetic energy will be lost by the rocket because of atmospheric resistance. When the rocket will reach its highest point the kinetic energy of the rocket will be zero and the potential energy will be equal to, Applying law of conservation of energy, |