Gravitation

If the interior contained matter is of same density everywhere, then we assume earth approximately as a point mass of mass equal to mass of earth, placed at the centre of earth. In that case, acceleration due to gravity (g) is zero at the centre of the earth. But it can’t be zero at any point on the surface of the earth.

The magnitude and direction of acceleration due to gravity can not be predicted at every point on the surface if the mass density is non - uniformly distributed everywhere. So, options(a), (b) and (c) are incorrect.

All the planets move around the sun in approximate circular orbits dur to gravitational force of attraction between the sun and the planet.

Where,

M₁ and M₂ are masses of two bodies and r is the distance between them

This gravitational force is inversely proportional to the square of distance between them and hence the planet revolves in circular orbits.

Similarly, Mercury experiences force of gravitational attraction both from the Sun and the Earth. But since the mass of sun is massive, the force due to sun is very large compared to force due to earth.

Therefore, mercury revolves in circular orbit around sun and not around earth.

if we consider earth and sun as a system, the earth experiences a centripetal force due to gravitational force of attraction of sun, and thus revolves around the sun in a circular orbit.

We know that,

Torque of any force F about a point is defined as

.(i)

Where,

r⃗ = distance of point of application of force from the point about which torque is to be calculated.

Since earth is symmetrical in shape, the net force due to sun acts on the centre of earth and since gravitational force is along line joining the centre of mass of earth and sun, the angle between r⃗ and F⃗ becomes 0°.

Therefore from eqn.(i)

τ = 0

Due to the viscous atmosphere, friction force due to the atmosphere acts on the satellite which reduces its orbital speed and hence the energy of revolution around a planet.

Due to decrease in energy of satellite, its height gradually decreases

The gravitational force of attraction due to earth on the moon follow the inverse square law due to which the as seen from the earth, the moon revolves around it in circular orbit.

When observed from the sun, moon experiences the gravitational pull due to both, the sun and the moon which results in a net force thus changing the trajectory of moon, and hence it does not revolve in strictly elliptical due to influence of both sun and earth.

Like any other planet, asteroids also follow Kepler’s law and move in elliptical orbits around planet due to the central gravitational force acting on them.

small mass of asteroid does not escape the gravitational force due to sun because of large mass of sun and hence, (a) is incorrect. Since gravitational force follow inverse square law, the asteroid will move in circular or elliptical orbits and (b) is incorrect option

Inertial mass:

The mass of an object A can be defined as the mass of the object obtained by measuring their accelerations under equal forces taking mass of B to be 1kg. This is known as inertial mass.

Gravitational mass:

The gravitational force exerted by a massive body on an object is proportional to the mass of the object. If F_A and F_B are the forces of attraction on the two objects due to the earth

and

Thus

If B is a standard unit mass, by measuring gravitational Force we can obtain the mass of object A which is known as gravitational mass.

By several experiments, it has been concluded that the inertial and gravitational mass are identical.

Therefore, gravitational mass of proton is equal to its inertial mass and is independent of the presence of neighbouring objects.

We know that gravitational force between two masses separated by a distance r is given by

Now,

Force due to 2M at A on B =

towards A

Force due to M at C on B =

towards C

Therefore, net force on B

Towards A

Therefore, m will move towards A

Therefore (c) is the correct option

When a particle is placed on the surface of the earth, then the acceleration due to gravity is given by

…(i)

Now when the particle is at a height h form the surface then new acceleration becomes

…..(ii)

From eqn. (i) and (ii)

Therefore, acceleration due to gravity decreases with increasing altitude

Now, if a body of mass m is placed on the point whose angle of latitude is ϕ then acceleration due to gravity is given by

Where,

ω = angular velocity of earth

R = radius of earth

As the latitude increases, ϕ increases and cosϕ decreases, thus increasing the value of g

Therefore, (a) and (c) are correct option

we know that, for planet to follow elliptical orbit

But if law of gravitation follows inverse cube law, then for a planet of mass m, the gravitational force experienced due to sun is given by

Orbital speed

The time period of revolution of planet is then given by

Which does not follow Kepler’s law. Hence, the planet will not revolve in either elliptical or circular orbit.

The new acceleration due to gravity can be given as

Since g’ is constant, hence path followed by the projectile will be approximately parabolic when thrown.

Given:

New gravitational constant G’ = 10G

M_s’ = M_s/10

Gravitational field due to earth is given by

Force on the man due to sun

Where r = distance between centre of sun and man

Since r≫R the effect due to sun can be neglected and the gravity pull will increase. Due to it, walking on the ground can be difficult

As the acceleration due to gravity g increases, the raindrops fall much faster than usual

Also, airplanes will have to travel much faster to overcome the increased gravitational pull of the earth

If the sun and the planets carried huge amounts of opposite charges then the electrostatic force of attraction will act between the sun and the planets.

We know that electrostatic or columbic force follows inverse square law and thus both the gravitational and electrostatic force will add up in nature and the net force will also follow inverse square law.

Therefore, since both the forces are of same nature, the Kepler’s law will still be valid

We know that gravitational force between the earth and the sun is given by

Where

G = Gravitational constant

M_e and M_s are mass of earth and sun respectively

Due to this force the earth revolves around the sun in circular orbits.

If the value of G decreases with time and the gravitational force between sun and earth starts decreasing and the trajectory of earth will change and hence not going around in strictly closed orbits.

As the gravitational pull of sun becomes weaker the radius of revolution of earth will gradually increase over time and after sufficiently long time, we will leave the solar system

Given:

Where

= constant

For calculating acceleration due to gravity let

M₁ = Mass of earth = M

M₂ = mass of object

= radius of earth = R

Then magnitude of force between the object and earth is given by

Which can be said as

Where K is a constant

Therefore, acceleration due to gravity can be calculated as

Which depends on mass and hence different for different objects

Therefore (a) is the correct option

since this force follows inverse square law, hence Kepler’s first two are valid but since g is no longer constant, therefore Kepler’s third law (Law of periods) does not remain valid

therefore (c) is the correct option

when n is negative than force is given by

Where a = - n

Therefore, force is inversely proportional to mass of body, hence an object lighter than water will experience more force than water and therefore will sink in water

Therefore (a), (c) and (d) are the correct options

Geostationary satellites are the satellites which revolve in a circular orbit around the earth in the equatorial plane with time period equal to the time period of rotation of earth = 24 hours. Hence the geostationary satellite appears fixed from any point on the earth.

Hence the geostationary satellite revolves with same speed and sense as of earth which is west - east direction.

Another kind of satellites known as polar satellites are low altitude of satellites which go around the poles of the earth in north - south direction. Its time period is 100 minutes.

Centre of mass of any object is described as mean position of mass in the object. Many objects with uniform shape and mass distribution have their centre of masses at their centre

Whereas centre of gravity of the object is the point where the gravity is assumed to act i.e. when the object is suspended from a fixed point, the torque of gravity about the point is zero.

Centre of mass and centre of gravity are not same points for an extended body with large dimensions due to different mass distribution. But if the size of the body is less and the gravitational field is uniform, then the centre of gravity and centre of mass are nearly at the same point.

In the case of apple falling from the tree, the mass of the apple is very large as compared to the air molecules underneath the apple. Thus, the vertical motion of apple dominates and due to gravitational force, it falls towards the earth.

Whereas an air molecule is surrounded by many air molecules having relatively equal mass. also due to random thermal motion of molecules their velocity is not directed downwards. Therefore, the air molecules do not fall towards earth. The density of air molecules close to the surface of earth is greater than other due to the gravitational force.

Central force:

Central force is a force which always directed at a fixed point either away or towards it and its magnitude depends upon the distance of body from the centre

Example: Gravitational force of attraction

Non - central force:

Non - central force is not directed towards or away from a fixed point.

Example: Magnetic force on a current carrying loop

According to Kepler’s second law, the line between the sun and the planet sweeps equal area in equal interval of time, i.e.

Where is the areal velocity of any planet and K is any constant

Therefore, areal velocity remains constant and is independent of time. The graph between areal velocity versus time for mars is as shown

We know that the areal velocity of earth around the sun is given by

….(i)

Where

L⃗ = angular momentum of earth about sun

mass of earth

Now, angular momentum of earth around the sun is given by

…. (ii)

Where

v⃗ = velocity of earth

r⃗ = position vector of earth w.r.t sun

from eqn.(i) and (ii) we have,

Therefore, direction of areal velocity is perpendicular to the plane containing velocity vector and position vector.

Therefore, areal velocity of earth is normal to the plane containing earth and the sun

The gravitational force between two - point masses is given by

Where, andare masses and r is the distance between them respectively & G is the universal gravitational constant.

This gravitational force is independent of the medium in which masses are placed since the value of G is a universal constant and independent of medium and the masses also remains unaffected.

Therefore, the gravitational force between the two - point masses when they are dipped in water keeping the separation between them same remains same when put in vacuum.

Yes, a body can have inertia but no weight. Inertia of a body is related to mass. Whereas weight of body is defined as

Where

M = mass of body

g = acceleration due to gravity

therefore, weight of the object can be zero when the acceleration due to gravity becomes zero.

Since we know that at the centre of earth, the value of acceleration due to gravity is zero and hence the weight of the object is zero at the centre, but it remains mass(inertia) remains constant.

Gravitational field between two bodies, unlike electric fields, does not depend upon the nature of the medium between the two bodies, therefore a body cannot be shielded by the influence of nearby matter by putting it inside a hollow sphere.

An astronaut in a small spaceship around earth is always in freefall and is constantly under the influence of gravitational force of earth and since the spaceship is small its gravitational force on the astronaut is very small as compared to earth, therefore the astronaut cannot detect gravity. If the spaceship is large enough to produce a comparable gravitational force against earth’s gravitational force the astronaut will then detect gravity.

Given:

Radius of hollow shell = R

Distance of point mass from centre of hollow shell = r

Let the mass of the spherical shell be M and the point mass be m. Then the force between the shell and the point mass is given by,

When point mass is positioned such that it is inside the hollow sphere i.e. 0<r<R, the force is zero as the mass of the hollow sphere is distributed on its surface. When the mass is positioned such that r≤R, the force will be,

Therefore, the graph will be as follows

Aphelion is the positon of earth from where it has the greatest distance from the sun and perihelion is the position from where it has the shortest distance. Now since the angular momentum of earth about the sun is conserved then we have:

Where m is the mass of the earth, r_ap is the earth - sun distance at the aphelion and r_pe is the earth - sun distance at the perihelion.

Since r_ap > r_pe, v_pe > v_ap, therefore the velocity at the perihelion is greater than the velocity at the aphelion.

The orbital plane of a satellite is the plane in which it revolves around the earth. The equatorial plane of earth is the plane which contains the equator.

(a) For a polar satellite, the satellite revolves from north to south and therefore its orbital plane is at 90° to the equatorial plane

(b) For a geostationary satellite whose position with respect to a place on earth is constant i.e. it revolves in the direction of the rotation of earth, the orbital plane is at 0° to the equatorial plane.

A day on earth is measured in terms of rotation of earth on its axis with respect to the sun called the mean solar day. When the day is measured with respect to a distant sun, it is called a sidereal day.

Let us look at the given figure, let the sun and distant star’s location be pointed by the long arrow at day 1, where they are directly in line with zenith of the observer on earth. When the earth rotates by 360° the arrow of the distant star points in the same direction along the zenith. However, the sun’s arrow is at an angle from the zenith. Thus, for the sun to be at the zenith in line with the observer, the earth needs to rotate an additional angle. Since the earth moves by 1° around the sun in a day, this additional angle is 1°. Now 360° is equal to 24 hours on earth or 1440 minutes, therefore

Therefore, the distant star will always rise 4 minutes earlier than the sun.

Given:

Distance between centres of spheres = 10 x radius

Let the mass of the two spheres be M and their radii be R. Then, the distance between the two will be 10R. let an object of mass m be placed at the midpoint of the line joining the centres of the spheres. Then the force acting on the object due to the two spheres F₁ and F₂ will be,

Therefore, the forces are equal and in opposite directions and therefore the mass is in stable equilibrium.

However, if the mass were to be displaced by say a distance x from the midpoint towards the sphere 2, the forces acting on the object would be

Therefore, F₂’ > F₁’ and therefore the net force acting on the mass will not be zero and equal to (F₂’ - F₁’) towards the sphere 2 and the object would be in unstable equilibrium.

Given

Radius of satellite’s orbit = R

(a) Let the mass of earth be M and the mass of the satellite be m. Then the orbital velocity of the satellite will be

then, the kinetic energy of the satellite will be

(b) the potential energy of the satellite is given by

(c) the total energy of the satellite is given by

Therefore, the curves are as follows

The trajectory traced by a projectile under the action of gravitational force will always be a conic section. When the distance thrown is small compared to the radius of the earth then the projectile follows a parabolic trajectory as demonstrated by the so - called projectile motion in the analysis of newton’s equations of motion. However, when the distance is comparable to the earth’s radius the trajectory is an ellipse according to Kepler’s laws. Therefore, the correct trajectory is (c)

Given:

Mass of the object = m

Radius of earth = R

Initial height of the object from earth’s centre = R

Final height of the object from earth’s centre = 2R

Let the mass of earth be M. We know that the gravitational potential energy of an object is given by,

Where r is the distance between the object and centre of earth. When the object is at its initial position it is at a distance of R from the centre of the earth. Then its potential energy,

When the object is raised to a height 2R, then the potential energy is given by,

Now the gain in gravitational potential energy is given by

Given:

Mass of the ring = M

Layout of the ring = r

Mass of the object = m

Initial distance of the object h = r

Final distance of the object 2h = 2r

To find the gravitational force on the object of mass m due to the ring let us consider an elemental mass on the ring of mass dM. The force due to this elemental mass is then,

where x is the distance of the object from the elemental mass m and

Now at the point P the force dF can be resolved into a vertical component and a horizontal component. If we consider two elemental masses diametrically opposite to each other, then the vertical component due to the two forces cancel each other as they are equal and opposite and the horizontal forces add up. Therefore, we only need to consider the horizontal component of the forces, therefore the net force on the object,

Now from the triangle OPA in the figure we have,

therefore,

When h = r

When h = 2h = 2r

Therefore, the force decreases by a factor of .

Given:

Radius of the orbit of planets = r

Let the mass of the sun be M and the planet be m.

(a) The force on an object revolving around the star is given by

Therefore, linear velocity v is inversely proportional to square root of r and hence when r increases, v decreases.

(b) The angular velocity ω is given by

Therefore, when r increases, ω decreases.

(c) The kinetic energy K is given by,

Therefore, as r increases, K decreases.

(d) The potential energy U is given by,

Therefore, as r increases U becomes less negative or increases.

(e) The total energy E is given by,

Therefore, as r increases E becomes less negative or increases.

(f) The angular momentum L is given by,

Therefore, as r increases, L increases.

Given

Side of hexagon = l

The force on the masses will be the resultant of the forces of all the other masses. Let the masses be m each. Now the distance AC is given by the parallelogram law i.e.

Due to symmetry we have AC = AE =

Now, for AD we have

Now the force on mass at A due to b is

Also

The force on mass A due to C is

Also due to symmetry, we have

The force on mass at A due to mass at D is

Now, the forces F_ae and F_ac make equal angles with the direction AD viz. 30°, and therefore by parallelogram law of vector addition their resultant is along AD and therefore the magnitude of the resultant,

Similarly, F_ab and F_af also make equal angles with AD viz. 60° and therefore they are also along AD and

Therefore, the net force on mass at A will be

(a) For a geostationary satellite, the centripetal force must equal the gravitational force. Let the mass of earth be M and the satellite be m, then

Where r is the radius of the orbit of the satellite and ω is its angular velocity

where T is the time period of revolution, then

Now the height h of the satellite from the surface of the earth,

where R is the radius of earth.

For a geostationary satellite, the time period is 24 hours or 86400 seconds,

(b) A geostationary satellite is visible form the equator in the region shown in the figure having an angle 2θ,

Now from the figure we have

Now 2θ = 162.58, therefore to cover the earth with such satellites so that from any point on the equator a satellite can be seen,

Therefore, we need a minimum of 3 satellites.

Given

The eccentricity of earth’s orbit e = 0.0167

Let the earth - sun distance at the aphelion and perihelion be r_a and r_p respectively, and the angular velocity at those points be ω_a and ω_p. Now since the angular momentum of the earth is conserved we have

Now, from the ellipse we have

Where a is the semi - major axis of the ellipse

Since e = 0.0167

Now let the mean angular velocity of earth corresponding to the mean solar day be ω, then

And since the mean angular velocity is the geometric mean of the two, we have

Now we know that the mean solar day corresponds to 1° on the orbit of earth. A mean solar day has 24 h for which the earth rotates 361°. Thus for aphelion the earth rotates 361.034° and hence the time for the day is 24 hours and 8.1 seconds, and the for the perihelion the earth rotates 359.96° and the time is 23 hours and 52 seconds. Thus the longest day is 8.1 seconds long and the shortest day is 8 seconds shorter. However, the variation of length of day in the year is not explained by this phenomenon as the total time of the day still remains around 24 hours, while the amount of time the sun is visible changes.

Given

Radius of earth = R = 6400 km

Aphelion earth - satellite distance r_a = 6R

Perihelion earth - satellite distance r_p = 2R

We know for an ellipse

Where e is the eccentricity of the ellipse, and the semi major axis.

Now, the angular momentum of the satellite is conserved we then, we have

Where m is the mass of satellite and v_p and v_a are the velocities at the perigee and apogee respectively

Now from conservation of energy at the apogee and perigee

Where K and U are the kinetic and gravitational potential energy of the satellite

Where M is the mass of earth. Using the relation between v_p and v_a above we have

Putting the values of r_p and r_a we have

To transfer the satellite to a circular orbit of radius 6R the satellite’s velocity has to be maintained at a value of