Gravitation

Gravitational force is always attractive, conservative and independent of medium and acts on line foining the centres of the bodies.

Newton's law of gravitation:

Force acting between two bodies directly proportional to product of the masses and inversly proportional to square of distance between them.

Given by,

$F=$$\frac{G{m}_1{m}_2}{r^2}$ where G= Universal gas constant=$6.67\ast 10^{\text{-11}}N{m}^{2}k{g}^{\text{-2}}$

Gravity is the phenomenon of attraction of surrounding bodies to centre of particular body and is a vector quantity It holds the atmosphere around earth.

Given by:

$F=mg$

$F=\frac{GMm}{r^2}$ so, $g=\frac{GM}{r^2}$

so, $g=\frac{4}{3}\pi G\rho r$

Variation of acceleration due to gravity

1.with height

$g^{\prime }={\left(\frac{R}{R+h}\right)}^2\ast g$

= $\left(1-\frac{2h}{R}\right)\ast g$ for h< < R

2.With depth

$g^{\prime }=\frac{GM}{R^3}\ast (R-x)$$=g\left(1-\frac{x}{R}\right)$

At center of earth g'=0; x=R

3.With rotation of earth

$g_{\text{eff}}=g-R{w}^{2}co{s}^2\theta$

At equator, $\theta =0°$ ; $g_{\text{eff}}=g-R{w}^2$

At poles, $\theta =90°$; $g_{\text{eff}}=g$ There is no effect of rotation at poles

Radius of earth= 6400 Km

Gravitational intensity: It is the force experienced by unit point mass placed at that point of consideration with gravitational field $V=\frac{w}{m}=-\frac{GM}{R}$. It is scalar quantity.

Gravitational Potential (V): It is work done to take a unit mass from infinity to a certain point of consideration within gravitational field.field $V=\frac{w}{m}=-\frac{GM}{R}$. It is scalar quantity.

Gravitational potential energy: It is work done to take a body of mass m from infinity to a point of consideration within gravitational field.

$V=-\frac{GmM}{r}$

Gravitational potential energy on height h from the surface of earth

$V=-\frac{GMm}{R+h}$

Its maximum value is 0 to infinity.

Change in gravitational PE when a body is raised from surface to height h is

$\Delta U=-\frac{GMm}{R}-\left(\frac{-GMm}{R+h}\right)$

=$mgh\left(\frac{R}{R+h}\right)$

Gravitational potential and intensity are related as

$E=-\frac{dV}{r}$

$V=\int -E.dr$

Gravitational field and potential

1.For point mass at distance r,

$E=\frac{GM}{r^2}$ and $V=-\frac{GM}{r}$

2.For hollow sphere of mass M having radius R

i. Outside point (r>R): $E=\frac{GM}{r^2}$ and $V=-\frac{GM}{r}$

ii. On the surface (r=R): $E=\frac{GM}{R^2}$ and $V=-\frac{GM}{R}$

iii. Inside point (r$E=0$

3.For solid sphere of mass M and radius R

i. Outside point (r>R): $E=\frac{GM}{r^2}$ and $V=-\frac{GM}{r}$

ii. On the surface (r=R): $E=\frac{GM}{R^2}$ and $V=-\frac{GM}{R}$

iii. Inside point (r$E=\frac{GM}{R^3}\ast r$

4.For ring of radius R, mass M on its axis

$E=\frac{GM}{(R^2+r^2{)}^{\frac{3}{2}}}\ast r$ $V=\frac{-GM}{(R^2+r^2{)}^{\frac{1}{2}}}$

#If two point masses $m_1$and $m_2$are kept at distance r then field intensity becomes zero at a distance

$\frac{G{m}_1}{x^2}=\frac{G{m}_2}{(r-x{)}^2}$ $\to$ $x=\frac{\sqrt{m_1}.r}{\sqrt{m_1}+\sqrt{m_2}}$

Motion of satellites

$\frac{GMm}{r^2}=\frac{m{v}^2}{r}$ $\to$ $v={\left(\frac{GM}{r}\right)}^{\frac{1}{2}}$

Time period

$T=\frac{2\pi r}{v}=\frac{2\pi }{\sqrt{GM}}\ast r^{\frac{3}{2}}$$\propto$ $r^{\frac{3}{2}}$

For nearer satellite, r=R so, $v={\left(\frac{G{R}^2}{R}\right)}^{\frac{1}{2}}=\sqrt{gR}$

and $T=2\pi {\left(\frac{R}{g}\right)}^{\frac{1}{2}}$= 84.6 min

Geostationary satellite appears stationary with respect to earth and launched on equateral plane (time period t = 24 hours) and height= 36,000 km

Energy of satellite

P.E. = $\frac{-GMm}{r}$

K.E.=$\frac{1}{2}m{v}^2=\frac{GMm}{2r}$= binding energy

T.E. = P.E. + K.E.= $\frac{-GMm}{2r}$= (PE=2TE=-2KE)

If a body is projected from earth then the maximum height attained by it is

$T{E}_{\text{initial}}=T{E}_{\text{final}}$

or, $\frac{1}{2}m{v}^2+\left(\frac{+GMm}{R}\right)=0+\left(\frac{-GMm}{R+h}\right)$

or, $\frac{v^2}{2}=\frac{hgR}{(R+h)}$

so, $h=\frac{R}{\left(\frac{2gR}{v^2}\right)-1}=\frac{R}{{\left(\frac{v_e}{v}\right)}^2-1}$

For escape velocity, h$\to$$\infty$ $v_e=\sqrt{2gR}$ called escape velocity = 11.2 km/s for each

$v_{\text{escape}}=\sqrt{2}\ast v_{\text{orbital}}$

So, on increasing the orbital velocity of satellite by 41.4% the satellite escapes ( or KE by 100%)