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## Karnataka 1st PUC Physics Question Bank Chapter 8 Gravitation

Introduction

Question 1.

Who recognised that all the bodies irrespective of their masses accelerate towards the centre of the earth with a constant acceleration?

Answer:

Galileo

Question 2.

Name the model proposed by Ptolemy for planetary motions in which all celestial bodies revolve around the earth.

Answer:

Geocentric model.

Question 3.

Name the solar model proposed by indian scientist Aryabhatta.

Answer:

Heliocentric model in which all the planets revolve around the sun.

Question 4.

Who proposed a definitive model in which the planets moved in circles around a fixed central sun?

Answer:

Nicolas Copernicus.

Kepler’s Laws

Question 1.

State and Kepler’s law of orbits.

Answer:

Kepler’s law of orbits: All planets move in elliptical orbits with the sun as one of the foci of the ellipse.

Figure shows an ellipse traced out by a planet around the sun. The closest point is P (perihelion) and the farthest point is A (aphelion).

Question 2.

State and explain Kepler’s law of areas.

Answer:

Kepler’s law of areas: The line that joins any planet to the sun sweeps equal areas in equal intervals of time. Planets move slower when they are farther from the sun that when they are nearer.

The planet P moves around the sun in an elliptical orbit. The shaded area is the area AA swept in a small interval of time ∆t.

Question 3.

State and explain Kepler’s law of periods.

Answer:

Kepler’s law of periods: The square of the time period of revolution of a planet is proportional to the cube of the semi-major axis of the ellipse traced by the planet.

According to the law of periods, if T is the period of revolution of the planet and a is semi-major axis of the elliptical path, then T^{2} ∝ a^{3}

If T_{1} and T_{2} are the periods of revolution of two planets revolving round the sun in elliptical orbots, and ai and a2 are the semi-major axes of their orbits, then

\(\frac{T_{1}^{2}}{T_{2}^{2}}=\frac{a_{1}^{3}}{a_{2}^{3}}\)

Question 4.

Which physical quantity is conserved in the case of law of area?

Answer:

“Angular momentum” is conserved in the case of law areas.

Question 5.

Show that the area swept by a planet in a given interval of time is constant.

Answer:

The law of areas is the consequence of the conservation of angular momentum which is valid for a central force. A central force is that force which acts along the straight line joining the two bodies. Let the sun be at the origin. Let the position and momentum of the planet denoted by \(\vec{r}\) and \(\vec{p}\) respectively. Then the area swept out by

the planet of mass m in time interval ∆t is ∆A which is given by

where \(\vec{v}\) is the velocity . \(\vec{L}\) is the angular momentum along \(\vec{r}\). \(\vec{L}\) is constant as the planet goes around. \(\frac{\Delta \vec{A}}{\Delta t}\) is a constant. This is a law of areas. Gravitation is a central force and hence the law areas follows.

Universal Law of Gravitation and Gravitational Constant

Question 1.

What is the term gravity?

Answer:

It is the force of attraction between the Earth and any object lying on or near the surface of Earth.

Question 2.

What is the term gravitation?

Answer:

It is the force of attraction between any two bodies in the universe. •

Question 3.

State Newton’s universal law of gravitation and define universal gravitational constant.

Answer:

Every body in the universe attracts every other body with a force which is directly proportional to the product of their masses and inversely proportional to the square of the distance between them.

Explanation: Consider two bodies of moves mi and m2 and separated by distance r.

According to the law,

where G is called the universal gravitational constant.

\(\frac{F r^{2}}{m_{1} m_{2}}\)

G = F, when r = 1 m m_{1} = m_{2} = 1 kg .

The universal gravitational constant is defined as the force of attraction between two bodies of unit mass each placed at arbitrary distance 1 m apart.

S.I. unit is N m^{2} kg^{-2}

G = 6.67 × 10^{-11}Nm^{2} kg^{-2}

dimensions of G = [M^{-1}L^{3}T^{-2}

Note: The gravitational force is attractive and the direction of the force -F is along r. According to Newton’s third law, the gravitational force on mi due to m2 is -F. Thus, the gravitational force F12 on the body 1 due to 2 and F2I on the body 2 due to 1 are related as F_{12} = – F_{21}.

Question 4.

What is the magnitude of force on a body of mass m placed at the centre (inside) of a hollow spherical shell of uniform density?

Answer:

Zero

Question 5.

What is the value of gravitational constant?

Answer:

G = 6.67 × 10^{-11} N m^{2} / kg^{2}

Question 6.

Express universal law of gravitation in vector form.

Answer:

The force \(\vec{F}\) on a point mass m_{2} due to another point mass m_{1} at a distance r is given by

\(\vec{F}=-\left[\frac{G\left(m_{1} m_{2}\right)}{|\vec{r}|^{3}}\right] \vec{r}\)

Question 7.

Why is ‘G’ called universal constant of gravitation?

Answer:

It is because, the value of G is independent of the medium between the two bodies, masses of two bodies and distance between two bodies. It remains the same in the entire universe.

Question 8.

Why does a body weights move at poles and cell at equator.

Answer:

As the value of g is move at poles than at the equator the body weighs move at the poles.

Question 9.

Does acceleration due to gravity depend upon the mass of a planet?

Answer:

Yew, it is because

g = \(\frac{G M}{R^{2}}\)

i.e., acceleration due to gravity is directly proportional to the mass of the planet.

Question 10.

Discuss the Cavendish’s experiment for the determination of value of G.

Answer:

The value of G was determined by Cavendish in 1798.

Consider two small lead spheres of mass m_{1} and m_{2} connected to the ends of a light rod of length L. The rod is suspended by a quartz fibre. Two large lead spheres of mass M each is brought close to the small spheres but in opposite direction as shown. The big spheres attract the small opposite forces as shown. It is observed that the small space moves towards the bigger sphere with a force F.

∴ F = \(\frac{G M_{E} m}{r^{2}}\)

The forces on the two small spheres due to the bigger spheres produces a couple which exerts a torque,

∴ Torque = force × perpendicular distance

t = F × L

Deflecting torque = \(\tau=\frac{G M_{E} m}{r^{2}} L\)

Restoring torque = K.θ.

At equilibrium

Deflecting torque = restoring torque

\(\frac{G M_{E} m}{r^{2}}\) L= K θ

Acceleration Due to Gravity of the Earth

Question 1.

What is meant by acceleration due to gravity of earth?

Answer:

The acceleration produced in a freely falling body under the gravitational pull of the earth is called acceleration due to gravity. It is a vector directed towards the centre of the earth.

Question 2.

What is meant by weight of the body?

Answer:

Weight of the body means the gravitational force with which a body is attracted towards the centre of the earth.

w-mg

w → weight of the body

m → Mass of the body

g → acceleration due to gravity.

Question 3.

Derive the relation between acceleration g due to gravity and gravitational constant G.

Answer:

Consider a body of mass m to be place on the surface of the earth of mass M and radius R.

The According to the law;

F = \(\frac{G M_{E} m}{R_{E}^{2}}\) …. (1)

This force F produces an acceleration g in the body of mass ‘m’, The according

to Newtons 2^{nd} Law;

F = mg … (2)

from (1) nd (2)

mg = \(\frac{G M_{E} m}{R_{E}^{2}}\)

g = \(\frac{G M_{E}}{R_{E}^{2}}\)

Acceleration Due to Gravity below and above the Surface of the Earth

Question 1.

Derive an Expression for acceleration due to gravity at a certain attitude or height. OR Show that the value of ‘g’ decreases with the altitude or height.

Answer:

Consider a body of mass m to the placed on the surface of the earth of Mass M and radius R. then,

F = \(\frac{G M_{E}}{R_{E}^{2}}\)

Let the body taken to a height h from the surface of the earth, then,

Question 2.

Derive an Expression for acceleration due to gravity at a certain depth, or Show that acceleration due to gravity decreases with depth, and becomes two at the centre of the earth.

Answer:

Consider a body of mass m to be placed on the surface of the earth of mass ME, radius RE and density p.

g = \(\frac{G M_{E}}{R_{E}^{2}}\)

M = volume × density

= \(\frac{4}{3} \pi R_{E}^{3} \times \rho\)

Consider another point at a depth ‘d’ below the surface of the earth then the radius becomes (R – d).

Question 3.

What is the acceleration due to gravity at the centre of the earth? ‘

Answer:

Acceleration due to gravity is zero at the centre of the earth.

Gravitational Potential Energy

Question 4.

What is meant by gravitational potential energy of a body?

Answer:

The gravitational potential energy of a body is the energy associated with it due to its position in the gravitational field of another body and is measured by the amount of work done in bringing a body from infinity to a given point in the gravitational field of the other.

Question 5.

Derive an Expression for gravitational potential energy of a body?

Answer:

Consider the Earth of Mass M_{E} and Radius R_{E}.

Let P be a point at a distance r from O.

To find potential energy at P, consider another point A at a distance x from O. Let a body of mass m is placed at A. Let this body is moved from A for B i.e., through a small distance dx.

The amount of work done in moving a body from A to B through a distance dx is given by; Workdone = force × displacement

dw = F × dx

dw = \(\frac{G M_{E} m}{x^{2}} \cdot d x\)

Total work done in moving a body of mass m from ∞ to the given point p is;

By definition this work done is the gravitational potential energy.

∴w = U

Escape Speed

Question 1.

Define escape velocity, and mention the expression for the same.

Answer:

Escape velocity is the minimum velocity with which a body must be projected vertically upwards in order that it just escape the gravitational influence of the earth.

V_{e} = \(\sqrt{2 g R_{E}}\)

V_{e} – escape velocity

g – acceleration due to gravity.

R_{E} – Radius of the earth.

Question 2.

Derive an Expression for escape velocity of an object from the surface of the earth.

Answer:

Consider the earth mass M and Radius R. Suppose a body of mass m lies at a point P at a distance x from its centre O, then gravitational force is given by, F = \(\frac{G M_{E} m}{x^{2}}\)

The amount of work done in moving the body through small distance dx i.e., from P to Q against the gravitational force is;

dw = Fdx = \(\frac{G M_{E} m}{x^{2}}\). dx

The total amount of work done in moving the body from the surface of the earth to a region beyond the gravitational field is;

If v_{e} is the escape velocity of the body, then kinetic energy is \(\frac{1}{2}\)mv_{e}^{2}

This \(\frac{1}{2}\)mv_{e}^{2} imparted to the body at the surface of the earth will be just sufficient to perform the work.

v_{e} = \(\sqrt{2 g R_{E}}\)

Question 3.

Does the escape velocity of an object depends on the location from where it is projected?

Answer:

Yes it depends on the location from where the object is projected.

Question 4.

What is the value of escape velocity of the object from the surface of earth?

Answer:

Question 5.

Define orbital velocity. Write the expression for the same.

Answer:

Orbital velocity is the velocity with which a satellite revolves round a planet.

V_{0} = \(\sqrt{g R_{E}}\)

g- acceleration due to gravity R_{E} – Radius of the earth.

Earth Satellites, Geostationary and Polar Satellites, Weightlessness

Question 1.

What is a satellite?

Answer:

A satellite is an object which revolves around the planet is circular or elliptical orbit.

Question 2.

What are earth satellites?

Answer:

Earth satellites are objects which revolve around the earth.

Question 3.

What are polar satellites?

Answer:

Polar satellites are low altitude (h = 500 to 800 km) satellites that go around the poles of the earth in a north- south direction.

Question 4.

What are geostationary satellites?

Answer:

Geostationary satellites are the satellites in circular orbits around the earth in the equatorial plane with the time period of 24 hours.

Question 5.

Give the period of geostationary satellites?

Answer:

The period of geostationary satellite is 24 hrs.

Question 6.

Name the group of the geostationary satellites launched by India?

Answer:

The group of the geostationary satellites sent up by India is INSAT group of satellites.

Question 7.

Give an important use of geostationary satellites.

Answer:

Geostationary satellites are widely used for telecommunications.

Question 8.

Define orbital speed of a satellite around the earth.

Answer:

Orbital speed of a satellite around the earth is the speed required to put a satellite into its orbit.

Question 9.

Find the Expression for the total energy of a satellite revolving around the surface of the earth.

Answer:

Potential energy which is given by,

when the satellite revolves close to the earth’s surface, h can be neglected.

Question 10.

Define time period of a satellite. Hence obtain the expression for period of a satellite.

Answer:

It is the time taken by a satellite to complete one revolution around the earth.

Question 11.

What is weightlessness?

Answer:

A body experiencing a free fall towards the centre of the earth has no ground reaction to support it. Under this condition, a body experiences weightlessness.

Numerical Problems

Question 1.

The planet Mars has two moons, phobos and delmos. (i) phobos has a period 7 hours, 39 minutes and an orbital radius of 9.4 × 103 km. Calculate the mass of mars, (ii) Assume that earth and mars move in circular orbits around the sun, with the martian orbit being 1.52 times the orbital radius of the earth. What is the length of the martian year in days?

Answer:

Where R_{Ms} is the mass – sun distance and _{RES} is the earth – sun distance.

∴ T_{M} = (1.52)^{3/2} × 365 = 684 days.

Question 2.

You are given the following data: g = 9.81 ms-2, RE = 6.37 × 106 m, the distance to the moon R = 3.84 × 10^{8} m and the time period of the moon’s revolution is 27.3 days. Obtain the mass of the Earth ME in two different ways.

Answer:

Question 3.

A 400 kg satellite is in a circular orbit of radius 2RE about the Earth. How much energy is required to

transfer it to a circular orbit of radius 4 RE? What are the changes in the kinetic and potential energies?

Answer:

The kinetic energy is reduced and it mimics ∆E, namely, ∆K = K_{f} – K_{i} = -3.13 × 10^{9} J

The change in potential energy is twice the change in t he total energy, namely

∆V = K_{f} – V_{i} = -6.25 × 10^{9} J = U_{f} – U_{i}

Question 4.

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

Answer:

For the satellite around Jupiter

T_{f} = 1.769 days, R_{f} = 4.22 × 10^{8} m For earth around sun

T_{e} = 365.25 days, R_{e} = 1.496 × 10^{11} m

We know that,

Thus mass of Sun is about thousand times mass of Jupiter.

Question 5.

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 10s ly.

Answer:

Number of stars in our galaxy = 2.5 × 10^{11}

Mass of each star – -1 solar mass = 2 × 10^{30}Mkg

Mass of our galaxy = 2.5 × 10^{11} ×2 × 10^{360}kg = 5 × 10^{41} kg

Diameter of milky way – 10^{5} ly

d = 10^{5} × 3 × 10^{8}ms^{-1} × 1yr

The distance of the star = radius of its orbit = 5 × 10^{4} ly.

Therefore, the time period of the revolving star is given by,

T = 3.549 × 10^{8}yrs

It takes about 3.55 × 10^{8} years to complete one revolution for a star around milky way.

Question 6.

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 × 10^{30} kg, mass of the earth = 6 × 10^{24} kg. Neglect the effect of other planets etc. (orbital radius = 1.5 × 10^{11} m).

Answer:

Let the distance from the earth to the point where gravitational force is to zero be x.

Given that,

Mass of sun = M_{s} = 2 × 10^{30} kg

Mass of Earth = M_{e} = 6 × 10^{24} kg

Orbital radius = r =1.5 × 10^{11} m

We know that, gravitational force ∝ \(\frac{M_{E}}{R^{2}}\)

Since, gravitational force of Earth = Gravitational force of Sun

At a distance of about 2.6 × 10^{5} km gravitational force is zero.

Question 7.

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:

Given that orbital radius, r= 1.5 × 10^{8} km= 1.5 × 10^{11} m We know that, Time for one revolution = 1 year = 365.25 × 24 × 60 × 60s

The estimate mass of sun is 2 × 10^{30} kg

Question 8.

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^{8} km away from the sun?

Answer:

Distance of earth from sun is the Earth’s orbital radius;

re =1.5 × 10^{8}km=1.5 × 10^{11}m

Time period of Earth = T_{e}

Time period of Saturn = (29.5 × T_{e})

From Kepler’s 3^{rd} Law of planetary motion

Question 9.

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?

Answer:

Weight of the body = 63 N = mg

Now at a certain height h, let the acceleration due to gravity be g_{n}.

So, New weight – mg_{n}

Question 10.

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?

Answer:

Weight of body = 250 N

Now, at a depth d, let acceleration due to gravity be gd,

Question 11.

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 × 10^{24} kg; mean radius of the earth = 6.4 × 10^{8} m; G = 6.67 × 10^{11} N m^{2} kg ^{-2}.

Answer:

Initial velocity of the rocket,

Vi- 5 × 103 ms^{-1}

Mass of the Earth, M_{E} = 6.0 × 10^{24} kg

Radius of the Earth, R_{E} = 6.4 × 10^{6} m

Let the rocket be of mass, ‘m’ and reach an height ‘h’

Now, the initial energy of the rocket is given by, Total initial energy (TE_{1}) = Initial Kinetic Energy (KE_{i}) + initial Potential Energy (PE_{i})

At the highest point, the kinetic energy is 0.

Total Energy at height ‘h’ (TE_{n}) = PE_{n}

The rocket goes 1.6 × 10^{6} m from surface of the Earth and (6.4 + 1.6) × 10^{6}

= 8.0 × 106 m from the center of the Earth.

Question 12.

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.

Answer:

Escape velocity of projectile = 11.2 km s^{-1}

v_{e} = 112 × 10^{4}ms^{-1}

Initial velocity of the projectile, v_{i} = 3v_{e}

= 3.36 × 10^{4}ms^{-1}

Let m be the mass of the projectile and vf be the final velocity after it has escaped from the Earth’s gravitational field. Initial Energy = KE_{i} + PE_{i}

Question 13.

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 x 1024 kg; radius of the earth = 6.4 × 10^{6} m; G = 6.67 × 10^{-11} N m^{2} kg^{-2}.

Answer:

Mass of the satellite = 200 kg.

Mass of the Earth = 6.0 × 10^{24} kg

Radius of Earth = 6.4 × 10^{6} m

Height of the satellite = 400 km

We know that orbital velocity of the satellite,

Question 14.

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:

Mass of the Stars, M_{E} = 2 × 10^{30} kg

Radius of the Stars, R = 10^{4} km = 10^{7} m

Distance between Stars, d = 10^{9} km = 10^{12} m

Initial velocity almost 0, So KE_{i} of each is 0. So, Initial potential energy

Question 15.

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:

Mass of each sphere =100 kg, Radius of each sphere = 0.1 m

At the mid point ME the gravitational pull by each sphere is exactly the same since they are identical spheres.

So the gravitational force is zero.

The potential M_{E} = \(\frac{G M_{E}}{(0.5)}-\frac{G M_{E}}{(0.5)}\)

= -4 GM_{E} = -4 × 6.67 × 10^{-11} × 100 = -2-668 x 10^{-8}J/kg

So, the potential at M is – 2.67 × 10^{-8} J/kg

An object at M will be in equilibrium.

Since net force on it is zero, the equilibrium is unstable as a slight change in position of the mass, the net force becomes non zero and it will move.

Question 16.

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 x 10^{24} kg, radius = 6400 km.

Answer:

Mass of the Earth = 6.0 × 10^{24} kg

Radius of Earth = 6.4 × 10^{8} m.

Height of the satellite from surface of Earth is 36000 km = 36000 km = 3.6 × 10^{7} m Distance of the satellite from centre of Earth is d = (3.6 × 10^{7} m + 6.4 × 10^{8} m) = 4.24 × 10^{7} m

Gravitational potential due to Earth on the satellite is v = \(-\frac{G M_{E}}{d}\)

= \(-\frac{6 \cdot 67 \times 10^{-11} \times 6 \times 10^{24}}{4 \cdot 24 \times 10^{7}}\)

= -9.438 10^{7} J/kg

Question 17.

A star 2.5 times the mass of the sun and collapsed to 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 x 10^{30} kg).

Answer:

Since the Star is rotating, there is a force pushing it outward (centrifugal force) and a force pulling it inwards.

(gravitational force).

If gravitational force (F_{G}) > Centrifugal force (F_{C}), then the object on equator remain stuck on the star.

Mass of Star M = 2.5(M_{sun})

= 2-5 × 2 × 10^{30}kg =5 × 10^{30}kg

Radius of star =12 km = 1.2 × 10^{4}m .

Where, r = radius of star =12 km

ω = angular speed = 2 × n × (1.2)

F_{c}= m × 1.2 × 10^{4} × (2.4π)^{2} = 6.82 × 10^{5} × mN

Since F_{C} < F_{G} the object is stuck to Star’s surface.

Question 18.

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 × 10^{8} km; G = 6.67 × 10^{-11} N m^{2} kg^{-2}.

Answer:

Mass of Spaceship, m_{s} = 1000 kg

Mass of mars M_{s} = 6.4 × 10^{23} kg;

Radius of mars r_{m} = 3.395 × 10^{6} m

Radius of the orbit of mars = 2.28 × 10^{8} km = 2.28 × 10^{11} m

Initial velocity of spaceship = 0

So, the total initial energy is only potential energy.

Question 19.

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? Mass of mars = 6.4 × 10^{23} kg; radius of mars = 3395 km; G = 6.67 × 10^{-11} N m^{2} kg^{-2}.

Answer:

Mass of Mars = 6.4 × 10^{23} kg

Radius of Mars, R = 3.395 × 10^{6} m.

Initial velocity of the rocket = 2 km s^{-1} = 2 × 10^{3} ms^{-1}

Let the mass of the rocket be m, then the initial kinetic energy = \(\frac{1}{2}\)mv^{2}

= \(\frac{1}{2}\)m × ( 2 × 10^{3})^{2} =2m × 10^{6}J

Since 20% of the energy is lost due to almost resistance, only 80% of energy is to be considered. Now,

So, the satellite reaches a height of 495 km.