Gravitation - Class 11 Engineering Physics - Extra Questions

Force of attraction between two masses,$$F=\dfrac{GMm}{{{r}^{2}}}$$

where,  $$r$$ is the distance from earth center.

$$ ma=\dfrac{G{{M}_{e}}m}{{{r}^{2}}} $$

$$ a=\dfrac{G{{M}_{e}}}{{{\left( h+{{R}_{e}} \right)}^{2}}} $$

Hence, acceleration above height h from earth surface is $$\dfrac{GM}{{{\left( h+{{R}_{e}} \right)}^{2}}}$$

1. If we consider only the  sun and the moon to be the determining factors, then No, the weight,  even though by a negligible amount, would not vary. But, there are a lot  of determining factors other than the Sun, and we know that there is no  gravitationally neutral point in the universe, as the distance would  have to be infinite, for in inverse square law to produce zero gravity.

2. The gravity of the sun is about 0.00589 m/s^2 at the radius of the earth's orbit, or about 0.0006 of the earth's gravity, g = 9.81 m/s^2

Gravitational energy is the energy that is possessed by the object in the gravitational field by the virtue of its position. The potential energy at a certain height is equal to the work required to bring the mass from infinity with constant velocity to that height.

The gravitational potential energy near a planet is given by $$U$$ such that:

$$U = \dfrac{{ - GmM}}{r}$$      ;

where M=Mass of planet

m=mass of body

r=distance of body from planet

Let height above earth surface be $$h\;{\rm{m}}$$. Then

$$g' = g\left( {1 - \dfrac{{2h}}{R_E}} \right)$$

$$\dfrac{{g'}}{g} = \left( {1 - \dfrac{{2h}}{R_E}} \right)$$

$$\dfrac{{0.2}}{{9.8}} = \left( {1 - \dfrac{{2h}}{R_E}} \right)$$

$$\dfrac{{2h}}{R_E} = 1 - \dfrac{1}{{49}}$$

$$\dfrac{{2h}}{R_E} = \dfrac{{48}}{{49}}$$

$$h = \dfrac{{48R_E}}{{98}}$$

$$= \dfrac{{24 \times 6.4 \times {{10}^6}}}{{49}}$$

$$ = 3.1347 \times {10^6}{\rm{m}}$$

$$ = 3134.7\;{\rm{km}}$$

So, at height of $$3134.7\;{\rm{km}}$$ value of $$g$$ is $$0.2\;{\rm{m/}}{{\rm{s}}^{\rm{2}}}$$

The Gravitational  Potential Energy at a distance R is,

$$U=\dfrac { GMm }{ r } $$

 And weight at distance 2r is

 $$W=\dfrac { GMm }{ { (2r) }^{ 2 } } =\dfrac { GMm }{ 4{ r }^{ 2 } } =\dfrac { U }{ 4r } $$

The energy possessed by an object because of its position in the gravitational field is called Gravitational potential energy. The most common application of gravitational potential energy can be applied to the objects that are over the Earth’s surface where the gravitational acceleration can be assumed to be constant at $$9.8\times m/sec^2$$

Given,

Weight= $$mg=250\,N$$
$$g=\dfrac{GM}{{{r}^{2}}}$$

$$g'=\dfrac{GM}{{{\left( 2r \right)}^{2}}}=\dfrac{1}{4}\dfrac{GM}{{{r}^{2}}}=\dfrac{g}{4}$$

Weight become, = $$mg'=\dfrac{mg}{4}=\dfrac{250}{4}=62.5\,N$$

Hence, weight at distance twice of earth radius is $$62.5\,N$$ 

When a satellite is orbiting earth, its velocity is its orbital velocity, given by 

Orbital velocity, $$V _ { e } = \sqrt { \dfrac { 2 G M } { r } }$$

Additional velocity required is

$$v = V _ { e } - V _ { o } = \sqrt { 2 g R } - \sqrt { g R } = \sqrt { g R } ( \sqrt { 2 } - 1 )$$

Given, that $$R=6400km=6.4\text{ }\!\!\times\!\!\text{ }{{10}^{6}}\,m$$

$$g=9.8\,m{{s}^{-2}}$$

$$v = \sqrt { g R } ( \sqrt { 2 } - 1 ) = \sqrt { 9.8 \times 6.4 \times 10 ^ { 6 } } ( \sqrt { 2 } - 1 ) = 3.28 \times 10 ^ { 3 } m / s$$

Hence, additional velocity required is $$3.28 \times 10 ^ { 3 } \mathrm { m } / \mathrm { s }$$

The value of acceleration due to gravity is

$${{g}_{1}}=\dfrac{GM}{R_{1}^{2}}............(1)$$

When radius gets shrink by 2% then, new radius becomes, $${{R}_{2}}=98$$% = 0.98, then value of acceleration due to gravity is

$${{g}_{2}}=\dfrac{GM}{R_{2}^{2}}=\dfrac{GM}{0.98\,{{R}_{1}}}..........(2)$$

Taking ratio of equations (1) and (2)

$$ \dfrac{{{g}_{1}}}{{{g}_{2}}}={{\left( \dfrac{{{R}_{2}}}{{{R}_{1}}} \right)}^{2}}={{\left( \dfrac{0.98{{R}_{1}}}{{{R}_{1}}} \right)}^{2}}=0.9604 $$

$$ \dfrac{{{g}_{1}}}{{{g}_{2}}}=0.9604 $$

$$ {{g}_{2}}=1.04\,{{g}_{1}} $$

Hence, g increases by 4%.

Mass of the bucket with the paint, $$m=0.6 kg$$

Height at which the bucket is placed, $$h=2 m$$

Potential energy of the bucket with the paint at the given height,

$$P.E.= mgh $$       

$$=0.6\times(9.8)\times2$$         

$$=11.76 J$$

P.E. on the floor$$=0$$

Loss in potential energy$$=11.76−0=11.76 J\approx 12 J$$

We know that,

Force = Mass $$ \times $$ Acceleration

$$ F = m \times a$$ 

$$ a = \dfrac{F}{m} $$  …..1

Where $$F$$ is the force on the object of mass m dropped from a distance r from the centre of earth of mass $$ M.$$

So, force exerted by the earth on the object is

$$ F = G \dfrac{M \times m}{r^2}$$ ...2 

$$ M $$ = Mass of Earth

$$ m $$  = mass of object

$$ r $$  = distance of object from centre of earth

Now, from equation 1 and 2,

$$ a = G \dfrac{M \times m}{r^2 \times m} $$ 

$$ a = G \dfrac{M}{r^2} $$ 

Now, from above,

$$ a = g = $$ Acceleration due to gravity

We also see that, although force is depending on the mass of the object,

$$ F = G \dfrac{M \times m}{r^2} $$ 

But acceleration due to gravity is independent of the mass.

$$ g = G \dfrac{M}{r^2} $$  

(i) Value of gravitational constant $$ (G)$$ 

(ii) Mass of Earth $$ (M) $$ 

(iii) Radius of Earth $$ (r) $$ 

As gravitational constant $$ G$$ and mass of earth $$M$$ are always constant, so the value of acceleration due to gravity $$g$$ is constant as long as the radius of earth remains constant.

At the surface of earth also, the value of $$g$$  is not constant.  $$ \left  ( g \propto \dfrac{1}{r^2}  \right  ) $$ . 

At the poles, radius of earth is minimum, hence the $$g$$ is maximum. Similarly at the equator, the radius of earth is maximum, and hence the value of $$g$$ is minimum. Also as we go up from the surface of the earth, distance from the centre of the earth increases and hence value of $$g$$ decreases.

The value of $$g$$ also decreases as we go inside the surface of earth and $$g$$ is zero at the centre of earth, as at the centre of the earth the object has mass around it, so net force cancels and thus net acceleration becomes zero.

Since, mass of an object is the amount of matter contained in the object and as we know that going deep inside the earth will not change the amount of matter contained in the object, so mass remains constant. 

In fact mass of an object always remains constant and does not change from one place to another.

On the other hand, weight of an object is the force with which earth pulls an object towards its center.

Let h be the height where the acceleration due to gravity is 1% less than on the surface of the earth. Then,

$$g' = g \left ( 1 - \dfrac{2h}{R} \right )$$ 

Here, g = 1% less than g.

$$\therefore g' = \dfrac{99}{100}g$$

$$\therefore \dfrac{99}{100}g = g \left ( 1 - \dfrac{2h}{R} \right )$$

$$0.99 = 1 - \dfrac{2h}{R}$$

$$\therefore \dfrac{2h}{R} = 0.01$$$$h = \dfrac{0.01 \times R}{2}$$

$$= \dfrac{0/01 \times 6400}{2}$$

$$= 32 km$$

The weight of a body placed at the centre of the earth is zero. This is because the mass at the exact cenrte of the earth is zero. Gravitational force exists only if there is mass in the object.

Let the height be $$'h'$$

Given,acceleration is one percent less, which is $$0.99g$$

$$\Rightarrow \quad \dfrac { 99 }{ 100 } g=\dfrac { GM }{ { \left( R+h \right)  }^{ 2 } } $$

As, $$ g=\dfrac { GM }{ { R }^{ 2 } } $$

$$\dfrac { 99 }{ 100 } ={ \left( 1+\dfrac { h }{ R }  \right)  }^{ -2 }$$

Using binomial rule, as $$h<<R$$

$$\dfrac { 99 }{ 100 } ={ \left( 1-\dfrac { 2h }{ R }  \right)  }$$

$$\Rightarrow \quad h=\dfrac { R }{ 200 } $$

Radius of earth is $$6400km$$

Therefore, $$h = 2400/200 = 32km$$

Calculating the acceleration due to gravity at depth R

Consider a spherical shell of radius r and thickness $$dr$$

Mass of this shell, $$ m = 4 \pi r^2 x \partial r$$

acceleration due to this mass at distance R from center = $$ \partial a = \dfrac {G 4 \pi r^2 x \partial r} { R^2}$$

$$ a_R = \int_{0}^{R} \dfrac {G 4 \pi r^2 x \partial r} { R^2} = \dfrac { 4 \pi x G R}{3}$$

Calculating acceleration due to gravity at surface

acceleration due to this mass at distance $$2$$R from center = $$ \partial a = \dfrac {G 4 \pi r^2 x \partial r} {4 R^2}$$

$$ a_{2R} = \int_{0}^{R} \dfrac {G 4 \pi r^2 x \partial r} { 4R^2} +  \int_{R}^{2R} \dfrac {G 4 \pi r^2 x \partial r} { 4R^2} = \dfrac { 4 \pi x G R}{4*3} + \dfrac { 4 \pi y G}{4 R^2} \left [ \dfrac{(2R)^2}{3} - \dfrac{R^3}{3} \right ]$$

Therefore:

$$\dfrac{a_R}{a_{2R}} = \dfrac{1}{1}$$

=> $$\dfrac{x}{y} = \dfrac{7}{3}$$

So, $$n = 7$$

The variation of$$g$$with height is $$g = g\left( {1 - \dfrac{{2h}}{r}} \right)$$

The variation of$$g$$with depth is $$g = g\left( {1 - \dfrac{d}{r}} \right)$$

Equating the above two equation we get the height as

$$h = \dfrac{d}{2}$$

$$ = 50km$$

By using Kepler’s law

$${T^2} \propto {r^3}$$

By substituting the time period and the value of radius we get

$${\left( {\dfrac{{{T_1}}}{{{T_2}}}} \right)^2} = {\left( {\dfrac{{{r_1}}}{{{r_2}}}} \right)^3}$$

$${\left( {\dfrac{{365}}{{{T_2}}}} \right)^2} = {\left( {\dfrac{{2{r_1}}}{{{r_1}}}} \right)^3}$$

$${T_2} = \dfrac{{365}}{{2\sqrt 2 }}$$

The time period will be$${\rm{129 days}}$$

The energy stored in an object due to its position above the Earth's surface is called gravitational potential energy. This energy is due to the force of gravity acting on an object.

$${E_P} = mgh$$

When a stone falls from a height then potential energy is converted to kinetic energy when it comes on the surface of earth.

The acceleration due to gravity varies with height as

$$g = g\left( {\dfrac{{{R^2}}}{{{{\left( {R + h} \right)}^2}}}} \right)$$

By substituting the values in the above given equation we get

$$g = 9.8\left( {\dfrac{{{{\left( {6.4 \times {{10}^6}} \right)}^2}}}{{{{\left( {6.4 \times {{10}^6} + 3.2 \times {{10}^6}} \right)}^2}}}} \right)$$

On solving the above equation we get the value of acceleration due to gravity at height of $$3200\;km$$

$$g = 4.4\;m/{s^2}$$

( a )  The acceleration due to gravity on the earth is given by

$$g = \dfrac {GM}{R^2} i.e. GM = R^2g$$  . . . . . . .( 1 )

The acceleration due to gravity at height 'h' from the surface of the earth is given by

$$g_h = \dfrac{GM}{(R+h)^2} i.e. GM = (R+h)^2 g_h$$ . . . . . . ( 2 )

From ( 1 ) and ( 2 ) we have,

$$\dfrac{g_h}{g} = \dfrac {R^2}{(R+h)^2}$$

$$\dfrac{g_h}{g} = (1 - \dfrac{2h}{R})$$  

( b )  The acceleration due to gravity on the surface of the earth is given by.

$$g = \dfrac{GM}{R^2}$$ . . . .( 1 )

let 'Q' be the density of the material of the earth.

 Now, mass = volume $$\times$$ density

 $$M = \dfrac{4}{3} \pi R^3 \times \rho$$

 Substituting in equation ( 1 ) we get 

 $$g = \dfrac{G}{R^2} \times \dfrac{4}{3} \pi R^3 \times \rho = \dfrac{4}{3} \pi GR \rho$$

 $$g = \dfrac{4}{3} \pi GR \rho$$ . . .. .. ( 2 )

 $$g_d = \dfrac{4}{3} \pi G(R-d) \rho$$  . . . .. .( 3 )

 Dividing equation( 3 ) by ( 2 ) we get 

 $$\dfrac{g_d}{g} = \dfrac{R-d}{R} = (1 - \dfrac{d}{R})$$

 $$g_d = g ( 1 - \dfrac{d}{R})$$