NEET Physics Gravitation Notes

NEET Physics Gravitation Notes

Kepler’s First Law Class 11

All planets revolve around the sun in an elliptical orbit, with the sun at one of the foci

A → perihelion; B → aphelion

Kepler’s Second Law

The imaginary line which connects the planet and the sun sweeps equal areas in equal intervals of time.

⇒i.e., area of velocity = \(\frac{\Delta \mathrm{A}}{\Delta \mathrm{t}}=\frac{\mathrm{L}}{2 \mathrm{~m}}=\text { constant }\)

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Note:

In the case of central force L is constant. (Gravitational force is a central force).

Kepler’s Third Law (Law Of Periods)

The period of revolution of a planet around the sun is directly proportional to the cube of its semimajor axis.

⇒ \(\mathrm{T}^2 \propto \mathrm{a}^3\)

Note:

If the average radius is given, then

⇒ \(\mathrm{T}^2 \propto \mathrm{r}^3\)

NEET Physics Gravitation Notes PDF Download

Universal Law of Gravitation

The force of attraction between two masses mx and m2 separated by a distance ‘r’ is given by

⇒ \(\mathrm{F}=\mathrm{G} \frac{\mathrm{m}_1 \mathrm{~m}_2}{\mathrm{r}^2}\)

Where G is known as the universal gravitational constant. G= 6.67×10^-11Nm²kg^-2

The law of gravitation is strictly valid for point masses.

Shell Theorems

1. A uniformly dense spherical shell exerts a force on a body situated outside the shell as if the entire mass of the shell is concentrated at its center.
2. The force of attraction due to a uniformly dense spherical shell of uniform density, on a point mass situated inside it is zero.

Acceleration due to gravity/Gravitational field intensity/Gravitational field

Gravitational force at a point is the force experienced by a unit mass when placed at that point.

⇒ \(g=\frac{F}{m}\)

Acceleration due to gravity on the surface of the earth:

⇒ \(\mathrm{g}=\frac{\mathrm{GM}_{\mathrm{E}}}{\mathrm{R}_{\mathrm{E}}^2}\)

Acceleration due to gravity due to a uniform solid sphere:

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⇒ \(\mathrm{g}_{\mathrm{p}}=\frac{\mathrm{GM}}{\mathrm{r}^2} ; \text { if } \mathrm{r} \geq \mathrm{R}\)

⇒ \(\mathrm{g}_{\mathrm{p}}=\frac{\mathrm{GM}}{\mathrm{R}^3} \mathrm{r} ; \text { if } \mathrm{r}<\mathrm{R}\)

Variation of ‘g’ with distance from the center of the sphere:

Best Notes for Gravitation NEET Preparation

Variation of acceleration due to gravity with height:

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

When h<<R,then

⇒ \(g(h)=g\left(1-\frac{2 h}{R}\right)\)

Where ‘g’ is acceleration due to gravity on the surface of earth.

Variation of acceleration due to gravity with depth:

⇒ \(g(d)=g\left(1-\frac{d}{R}\right)\)

Variation of ‘g’ with latitude due to rotation of earth:

NEET Previous Year Questions on Gravitation

Consider a body of mass ‘m’ at point P.

w.k.t.,

⇒ \(g=\frac{F_{\text {bet }}}{m}\)

⇒ \(g_\lambda=\frac{m g-m \omega^2 R \cos ^2 \lambda}{m}\)

⇒ \(g_\lambda=g-\omega^2 R \cos ^2 \lambda\)

Case 1:

At equator, λ = 0

⇒ \(g_\lambda=g-\omega^2 R\)

If ‘ ω’ of earth increases there is a possibility of ‘g’ at equator becoming zero.

i.e., 0 = g – ω²R ⇒ g=ω²R

⇒ \(\omega=\sqrt{\frac{g}{R}}\)

,i.e., when ω=\(\sqrt{\frac{g}{R}}\) g at equator will become zero.

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Case 2:

At pole,  λ = 90°,

∴ g_λ = 0

i.e., ‘g’ at poles is independent of rotation of earth.

Gravitational potential energy at a point is the work done in bringing a mass from infinity to that point against the gravitational force of the field.

⇒ \(\mathrm{U}=-\frac{\mathrm{Gm}_1 \mathrm{~m}_2}{\mathrm{r}}\)

Where ‘r’ is the distance from the centre of earth. For system of ‘3’ particles,

⇒ \(\mathrm{U}=-\mathrm{G}\left(\frac{\mathrm{m}_1 \mathrm{~m}_2}{\mathrm{r}_{12}}+\frac{\mathrm{m}_2 \mathrm{~m}_3}{\mathrm{r}_{23}}+\frac{\mathrm{m}_3 \mathrm{~m}_1}{\mathrm{r}_{31}}\right)\)

For ‘n’ particles,

⇒ \(\mathrm{U}=-\mathrm{G}\left(\frac{\mathrm{m}_1 \mathrm{~m}_2}{\mathrm{r}_{12}}+\frac{\mathrm{m}_1 \mathrm{~m}_3}{\mathrm{r}_{13}}+\ldots . .+\frac{\mathrm{m}_1 \mathrm{~m}_{\mathrm{n}}}{\mathrm{r}_{\mathrm{ln}}}+\frac{\mathrm{m}_2 \mathrm{~m}_3}{\mathrm{r}_{23}}+\ldots . .\right)\)

Note:

For‘n’particle system, there are \({ }^{\mathrm{n}} \mathrm{C}_2=\frac{\mathrm{n}(\mathrm{n}-1)}{2}\) pairs.

Potential energy is calculated for each pair and then added to obtain the potential energy for the system.

⇒ \(F=-\frac{d U}{d x}\)

∴ \(U=-\int_{\infty}^r \vec{F} \cdot d \vec{x}\)

Gravitational potential at a point is the work done in bringing a unit mass from infinity to the given point against the gravitational force of the field.

⇒ \(\mathrm{V}=-\frac{\mathrm{GM}}{\mathrm{r}}\)

Newton’s Law of Gravitation NEET MCQs with Answers

Relation between gravitational field and gravitational potential

⇒ \(\mathrm{g}=-\frac{\mathrm{dV}}{\mathrm{dr}}\)

⇒ \(\mathrm{V}=-\int_{\infty}^{\mathrm{r}} \overrightarrow{\mathrm{g}} \cdot \mathrm{d} \overrightarrow{\mathrm{r}}\)

Potential due to solid sphere (if r > R )

⇒ \(V_p=-\int_{\infty}^r-\frac{G M}{r^2} \cdot d r\)

⇒ \(\mathrm{V}_{\mathrm{p}}=+\mathrm{GM}\left[\frac{\mathrm{r}^{-2+1}}{-2+1}\right]_{\infty}^{\mathrm{r}}=-\frac{\mathrm{GM}}{\mathrm{r}}\)

Potential due to solid sphere (if r < R)

NCERT Summary of Gravitation for NEET Physics

⇒ \(V_P=-\int_{\infty}^R-\frac{G M}{r^2} \cdot d r-\int_R^r-\frac{G M}{R^3} r d r\)

⇒ \(\mathrm{V}_{\mathrm{P}}=\mathrm{GM}\left[\frac{\mathrm{r}^{-1}}{-1}\right]_{\infty}^{\mathbb{R}}+\frac{\mathrm{GM}}{\mathrm{R}^3}\left[\frac{\mathrm{r}^{+2}}{+2}\right]_{\mathbb{R}}^{\mathrm{r}}\)

⇒ \(V_p=-\frac{G M}{R}+\frac{G M}{2 R^3}\left(r^2-R^2\right)\)

⇒ \(V_p=\frac{G M}{2 R^3}\left(r^2-R^2\right)-\frac{G M}{R}\)

⇒ \(V_P=\frac{G M}{2 R^3}\left[\left(r^2-R^2\right)-2 R^2\right]\)

⇒ \(V_P=-\frac{G M}{2 R^3}\left[3 R^2-r^2\right]\)

Orbital Speed of a Satellite

When gravitational force and centrifugal force on the satellite are same, then

⇒ \(\frac{\mathrm{GMm}}{\mathrm{r}^2}=\frac{\mathrm{mv}_0^2}{\mathrm{r}}\)

⇒ \(\mathrm{v}_0^2=\frac{\mathrm{GM}}{\mathrm{r}}\)

⇒ \(v_0=\sqrt{\frac{G M}{r}}=\sqrt{\frac{G M}{R+h}}\)

Where ‘h’ is the height of the satellite from earth’s surface. If h < < R, then,

⇒ \(\mathrm{v}_0=\sqrt{\frac{\mathrm{gR}^2}{\mathrm{R}}} \Rightarrow \mathrm{v}_0=\sqrt{\mathrm{gR}}\)

Step-by-Step Solutions for Gravitation NEET Problems

Expression for escape speed

⇒ \(v_e=\sqrt{\frac{2 G M}{r}}=\sqrt{\frac{2 G M}{R+h}}\)

If h < < R, then

⇒ \(\mathrm{v}_{\mathrm{e}}=\sqrt{\frac{2 \mathrm{gR}^2}{\mathrm{R}}} \Rightarrow \mathrm{V}_{\mathrm{e}}=\sqrt{2 \mathrm{gR}}\)

Note:

If a satellite is very close to the earth’s surface, then,

⇒ \(\mathrm{v}_{\mathrm{e}}=\sqrt{2} \mathrm{v}_0\)

Gravitational Potential Energy and Escape Velocity NEET Notes

Total Energy of a Satellite

The kinetic energy of a satellite is given by,

⇒ \(\mathrm{K}=\frac{1}{2} \mathrm{mv}_0^2=\frac{1}{2} \mathrm{~m} \frac{\mathrm{GM}}{\mathrm{r}}=\frac{\mathrm{GMm}}{2 \mathrm{r}}\)

Potential energy of a satellite is given by,

⇒ \(\mathrm{U}=-\frac{\mathrm{GMm}}{\mathrm{r}}\)

∴ Total energy,

⇒ \(\mathrm{E}=\mathrm{K}+\mathrm{U}=\frac{\mathrm{GMm}}{2 \mathrm{r}}-\frac{\mathrm{GMm}}{\mathrm{r}}\)

⇒ \(\mathrm{E}=-\frac{\mathrm{GMm}}{2 \mathrm{r}}\)