<span style="color: #000000;"><strong><span style=""class="fontSize14">Universal Law of Gravitation</span></strong></span></br>
<div><span style=" color: #000000;"class="fontSize10">$$F \propto \dfrac{m_1 m_2}{ r^2}$$ or $$F=G \dfrac{m_1 m_2 }{r^2}$$</span></div>
<div><span style=" color: #000000;"class="fontSize10">where $$G=6.67 \times 10^{-11} \ Nm^2 kg^{-2}$$ is the universal gravitational constant.</span></div>
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<span style="color: #000000;"><strong>Newton's Law of Gravitation in vector form :</strong></span></br>
<div><span style=" color: #000000;"class="fontSize10">$$\vec{F_{12}}= \dfrac{G m_1 m_2}{r^2} \hat{r_{12}}$$ and  $$\vec{F_{21}}= \dfrac{G m_1 m_2}{r^2} \hat{r_{21}}$$</span></div>

<div><span style=" color: #000000;"class="fontSize10">Now, $$\hat{r}_{12}=- \hat{r}_{21}$$, Thus $$\vec{F}_{21}= \dfrac{-G m_1 m_2}{r^2} \hat{r}_{12}$$</span></div>
<div><span style=" color: #000000;"class="fontSize10">Comparing above, we get $$\vec{F}_{12}=- \vec{F}_{21}$$</span></div>

<span style=" color: #000000;"class="fontSize14"><strong>Gravitational Field</strong></span></br>

<span style=" color: #000000;"class="fontSize10">Gravitational field across mass $$m$$ due to the mass $$M$$, $$E=\dfrac{F}{m}=\dfrac{GM}{r^2}$$</span></br>

<span style="color: #000000;"><strong>Gravitational potential:</strong></span></br>

<span style=" color: #000000;"class="fontSize10">$$V=\dfrac{-GM}{r}$$ and $$E=\dfrac{-dV}{dr}$$</span></br>

<span style=" color: #000000;"class="fontSize10"><strong>Ring</strong></span></br>

<span style=" color: #000000;"class="fontSize10">$$V=\dfrac{-G M}{ (a^2 + r^2)^{1/2} }$$ and $$E=\dfrac{-G M}{ (a^2 + r^2)^{3/2} } \hat{r}$$ or $$E=\dfrac{- G Mcos \theta}{x^2}$$</span></br>

<div><span style=" color: #000000;"class="fontSize10">Gravitational field is maximum at a distance, $$r= \pm a/ \sqrt2$$ and it is $$E=\dfrac{- GM}{ 3 \sqrt3 a^2}$$</span></div>
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<span style=" color: #000000;"class="fontSize12"><strong>Thin Circular Disc</strong></span></br>
<div><span style=" color: #000000;"class="fontSize10">$$V=\dfrac{-2 GM}{a^2} \left( \left(a^2 + r^2 \right)^{1/2} -r \right)$$ and $$E= \dfrac{-2 GM}{a^2} \left( 1- \dfrac{r}{ \left( r^2 + a^2 \right)^{1/2}} \right) =\dfrac{-2GM}{a^2} \left( 1-cos \theta \right)$$</span></div>

<span style="color: #000000;"><strong><span style=""class="fontSize12">Non conducting solid sphere</span></strong></span></br>
<div><span style=" color: #000000;"class="fontSize10"><b>(a) Point P inside the sphere. </b>$$r \leq a$$, then</span></div>
<div><span style=" color: #000000;"class="fontSize10">$$V=\dfrac{- GM}{2 a^3} ( 3a^2 - r^2) $$ and $$E= \dfrac{-G M r}{ a^3}$$, and at the centre $$V=\dfrac{- 3GM}{ 2a}$$ and $$E=0$$</span></div>
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<div><span style=" color: #000000;"class="fontSize10"><b>(b) Point P outside the sphere. </b>$$r \geq a$$, then</span></div>
<div><span style=" color: #000000;"class="fontSize10">$$V=\dfrac{-GM}{r}$$ and $$E= \dfrac{-G M}{ r^2}$$</span></div>

<span style="color: #000000;"><strong><span style=""class="fontSize12">Uniform Thin Spherical Shell / Conducting solid sphere</span></strong></span></br>
<div><span style=" color: #000000;"class="fontSize10"><b>(a) Point P Inside the shell. </b>$$r \leq a$$, then</span></div>
<div><span style=" color: #000000;"class="fontSize10">$$V =\dfrac{-G M}{a}$$ and $$E=0$$</span></div>
<div><span style=" color: #000000;"class="fontSize10"><b>(b) Point P outside shell. </b>$$r \geq a$$, then</span></div>
<div><span style=" color: #000000;"class="fontSize10">$$ V=\dfrac{-GM}{r}$$ and $$E= \dfrac{-G M}{ r^2}$$</span></div>

<span style=" color: #000000;"class="fontSize14"><strong>VARIATION OF ACCELERATION DUE TO GRAVITY:</strong></span></br>

<span style="color: #000000; "class="fontSize12"><strong>Effect of Altitude</strong></span></br>
<div><span style=" color: #000000;"class="fontSize10">$$g_h =\dfrac{G M_e}{ (R_e + h)^2} = g \left( 1 + \dfrac{h}{ R_e} \right)^{-2} \approx g \left( 1 - \dfrac{2h}{ R_e} \right)$$ when $$h\ll R$$</span></div>

<span style="color: #000000;"><strong><span style=""class="fontSize12">Effect of depth</span></strong></span></br>
<div><span style=" color: #000000;"class="fontSize10">$$g_d =g \left( 1 - \dfrac{d}{ R_e} \right)$$</span></div>

<span style=" color: #000000;"class="fontSize12"><strong>Effect of the surface of Earth</strong></span></br>
<div><span style=" color: #000000;"class="fontSize10">The equatorial radius is about 21 km longer than its polar radius. We know, $$g =\dfrac{ G M_e}{ R_e ^2}$$. Hence, $$g_{pole} > g_{equator}$$</span></div>
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<div><span style=" color: #000000;"class="fontSize10">Here,</span></div>
<div><span style=" color: #000000;"class="fontSize10">$$M_e$$ is the mass of earth.</span></div>
<div><span style=" color: #000000;"class="fontSize10">$$R_e$$ is the radius of earth.</span></div>
<div><span style=" color: #000000;"class="fontSize10">$$h$$ is the height above the earth surface.</span></div>
<div><span style=" color: #000000;"class="fontSize10">$$d$$ is the depth below the earth surface.</span></div>
<div><span style=" color: #000000;"class="fontSize10">$$G$$ is the gravitational constant.</span></div>
<div><span style=" color: #000000;"class="fontSize10">$$g$$ is the acceleration due to gravity. $$(g =9.81 \ ms^{-2})$$</span></div>
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<span style=" color: #000000;"class="fontSize10"><strong>Variation of g due to Rotation of Earth</strong></span>

<div><span style=" color: #000000;"class="fontSize10">$$g'=g- Rw^2 cos^2 \theta$$</span></div>
<div><span style=" color: #000000;"class="fontSize10">At a latitudinal position $$\theta$$ on earth, where $$w$$ is rotational velocity of the earth, the variation on gravitational acceleration follow above relationship.</span></div>

<span style=" color: #000000;"class="fontSize14"><strong>SATELLITE VELOCITY (OR ORBITAL VELOCITY)</strong></span></br>
<div><span style=" color: #000000;"class="fontSize10">$$v_o =\left( \dfrac{G M_e}{ R_e + h} \right)^{1/2} = \left( \dfrac{g R_e ^2}{ R_e + h} \right)^{1/2}$$</span></div>
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<div><span style=" color: #000000;"class="fontSize10">when $$h<< R_e$$, the $$v_o = \sqrt{g R_e}$$</span></div>
<div><span style=" color: #000000;"class="fontSize10">$$\therefore v_o = \sqrt { 9.8 \times 6.4 \times 10^6} = 7.92 \times 10^3 \ ms^{-1} =7.92 \ kms^{-1}$$</span></div>
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<span style=" color: #000000;"class="fontSize12"><strong>Time period of Satellite</strong></span></br>
<div><span style=" color: #000000;"class="fontSize10">$$T = \dfrac{ 2 \pi (R_e + h) }{ \left( \dfrac{g R_e ^2}{ (R_e + h) } \right)^{1/2} } = \dfrac{2 \pi}{ R_e} \left( \dfrac{(R_e + h)^3}{g} \right)^{1/2}$$</span></div>
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<span style=" color: #000000;"class="fontSize12"><strong>Energy of a Satellite</strong></span></br>
<div><span style=" color: #000000;"class="fontSize10">$$U =\dfrac{ - GM_e m}{ r} $$ $$K.E. = \dfrac{ G M_e m}{ 2r}$$ ; then total energy $$T.E = P.E + K.E = \dfrac{ G M_e m}{ 2r}$$</span></div>

<span style=" color: #000000;"class="fontSize14"><strong>Kepler's Laws</strong></span></br>

<span style=" color: #000000;"class="fontSize12"><strong>Law of area :</strong></span></br>

<span style=" color: #000000;"class="fontSize10">The line joining the sun and a planet sweeps out equal areas in equal intervals of time.</span></br>

<span style=" color: #000000;"class="fontSize10">$$\text{ Areal velocity} = \dfrac{ \text{ area swept}}{ \text{ time}} = \dfrac{ \frac{1}{2} r (r d \theta) }{dt} =\dfrac{1}{2} r^2 \dfrac{d \theta}{ dt} = Constant$$</span></br>

<span style=" color: #000000;"class="fontSize10">Hence,</span></br>

<span style=" color: #000000;"class="fontSize10"><b>Law of periods </b>: $$\dfrac{ T^2}{ R^3} = Constant$$</span></br>