Here students can locate TS Inter 1st Year Physics Notes 9th Lesson Gravitation to prepare for their exam.
TS Inter 1st Year Physics Notes 9th Lesson Gravitation
→ Kepler’s Laws :
Law of orbits (1st law) ‘.All planets move in an elliptical orbit with the sun is at one of its foci.
→ Law of areas (2nd law) : The line joining the planet to the sun sweeps equal areas in equal intervals of time, i.e., \(\) = constant.
i. e., planets will appear to move slowly when they are away from sun, and they will move fast when they are nearer to the sun.
→ Law of periods (3rd law) : The square of time period of revolution of a planet is proportional to the cube of the semi major axis of the ellipse traced out by the planet.
i.e., T2 ∝ R3 ⇒ \(\frac{\mathrm{T}^2}{\mathrm{R}^3}\) = constant
→ Newton’s law of gravitation (OR) Universal law of gravitation: Every body in universe attracts other body with a force which is directly proportional to the product of their masses and inversely proportional to the square of the distance between them.
F ∝ m1m2, F ∝ \(\frac{1}{\mathrm{r}^2}\) ⇒ F = G\(\frac{\mathrm{m}_1 \mathrm{~m}_2}{\mathrm{r}^2}\)
→ Central force : A central force is that force which acts along the line joining the sun and the planet or along the line joining the two mass particles.
→ Conservative force : For a conservative force work done is independent of the path. Work done depends only on initial and final positions only.
→ Gravitational potential energy : Potential energy arising out of gravitational force is called gravitational potential energy.
Since gravitational force is a conservative force gravitational potential depends on position of object.
V = \(-\frac{\mathrm{Gm}_1 \mathrm{~m}_2}{\mathrm{r}}\)
→ Gravitational potential : Gravitational potential due to gravitational force of earth is defined as the “potential energy of a particle of unit mass at that point”.
Gravitational potential V = \(\frac{G M}{r}\)
(r = distance from centre of earth)
→ Acceleration due to gravity (g) :
Acceleration due to gravity ‘g’ = \(\)
→ Acceleration due to gravity below and above surface of earth :
1) For points above earth total mass of earth seems to be concentrated at centre of earth.
For a height ‘h’ above earth
g(h) = \(\frac{\mathrm{GM}_{\mathrm{E}}}{\left(\mathrm{R}_{\mathrm{E}}+\mathrm{h}\right)^2}\)
where h << RE
g(h) = g\(\left(1+\frac{h}{R_E}\right)^{-2}\) = g\(\left(1-\frac{2 \mathrm{~h}}{\mathrm{R}_{\mathrm{E}}}\right)\)
2) For a point inside earth at a depth’d’ below the ground mass of earth (Ms) with radius (RE – d) is considered. That mass seems to be at centre of earth.
g’ = g\(\left(1-\frac{d}{R}\right)\)
→ Escape speed (v1)min : The minimum initial velocity on surface of earth to overcome gravitational potential energy is defined as “escape speed ve”
ve = \(\sqrt{2 \mathrm{gR}}=\sqrt{\frac{2 \mathrm{GM}}{\mathrm{R}}}\)
→ Orbital velocity: Velocity of a body revolving in the orbit is called orbital velocity.
Orbital velocity V0 = \(\sqrt{\frac{\mathrm{GM}}{\mathrm{R}}}\) ⇒ V0 = \(\sqrt{\mathrm{gR}}\)
Note:
- Relation between orbital velocity V and escape speed ve = √2V0
- If velocity of a satellite in the orbit is increased by √2 times or more it will go to infinite distance.
→ Time period of the orbit (T) : Time taken by a satellite to complete one rotation in the orbit is called “time period of rotation”.
T = 2π\(\frac{\left(\mathrm{R}_{\mathrm{E}}+\mathrm{h}\right)^{3 / 2}}{\sqrt{\mathrm{GM}_{\mathrm{E}}}}\)
→ Geostationary orbit : For a geostationary orbit in equatorial plane its time period of rotation is 24 hours, i.e., angular velocity of satellite in that orbit is equal to angular velocity of rotation of earth.
Geostationary orbit is at a height of 35800 km from earth.
→ Geostationary satellite : Geostationary satellite will revolve above earth in geostationary orbit along the direction of rotation of earth. So it always seems to be stationary w.r.t earth.
Time period of geostationary satellite is 24 hours. It rotates in equatorial plane in west to east direction.
→ Polar satellites: Polar satellites are low attitude satellites with an altitude of 500 km to 800 km. They will revolve in north-south direction of earth.
Time period of polar satellites is nearly 100 minutes.
→ Weightlessness: Fora freely falling body its weight seems to be zero. Weight of a body falling downwards with acceleration ‘a’ is w’ = mg’ = m(g – a). When a = g the body is said to be under free fall and it seems to be weightless.
→ Force between two mass particles, F = \(\frac{\mathrm{Gm}_1 \mathrm{~m}_2}{\mathrm{r}^2}\)
→ Universal gravitational constant, G = \(\frac{\mathrm{Fr}^2}{\mathrm{~m}_1 \mathrm{~m}_2}\)
G = 6.67 × 10-11 Nm2 / Kg2 D.F.: M-1 L3 T-2
→ Relation between g and G is,
g = \(\frac{\mathrm{GM}}{\mathrm{R}^2}=\frac{4}{3}\)πρG.R
→ Variation ofg with depth, gd = g(1 – \(\frac{d}{R}\))
→ Variation ofg with height, gh = g(1 – \(\frac{2h}{R}\))
For small values of h i.e., h < < R then
gh = g(1 – \(\frac{2h}{R}\))
(a) Gravitational potential, U = –\(\frac{\mathrm{GMm}}{\mathrm{R}}\)
(b) If a body is taken to a height h’ above the ground then
Gravitational potential, Uh = –\(\frac{\mathrm{GMm}}{(\mathrm{R}+\mathrm{h})}\)
→ Orbital velocity, V0 = \(\sqrt{\frac{\mathrm{GM}}{\mathrm{R}}}=\sqrt{\mathrm{gR}}\)
→ Orbital angular velocity, ω0 = \(\sqrt{\frac{\mathrm{GM}}{\mathrm{R}^3}}=\sqrt{\frac{\mathrm{g}}{\mathrm{R}}}\)
→ Escape velocity, Ve = \(\sqrt{\frac{2 \mathrm{GM}}{\mathrm{R}}}=\sqrt{2 \mathrm{gR}}\)
→ Time period of geostationary orbit = 24 hours.
→ Angular velocity of earth’s rotation
(ω) = \(\frac{2 \pi}{24 \times 60 \times 60}\) = 0 072 × 10-3 rad/sec