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![ W = -GMm [
1
𝑅
-
1
∞
]
U = -
𝐺𝑀𝑚
𝑟
Gravitational Potential
Amount of work done in bringing a body of unit mass from infinity to the given
point
Gravitational Potential = -
𝐺𝑀
𝑟](https://image.slidesharecdn.com/gravitationppt-180122143106/85/GRAVITATION-CLASS-11TH-17-320.jpg)
![ Proof :
𝜏 = 𝑟→ x 𝐹→ = 0 [ opp. Faced]
𝜏 =
𝑑𝐿→
𝑑𝑡
= 0.
Or 𝐿→ = constant .
∆𝐴
→
= 𝐴𝑟𝑒𝑎 𝑜𝑓 𝑡ℎ𝑒 𝑡𝑟𝑖𝑎𝑛𝑔𝑙𝑒
∆𝐴
→
=
1
2
𝑟→
x 𝑃→
P (∆𝐴
→
= Area)
𝑃→
P = ∆ 𝑟→
= 𝑣→
∆t =
𝑃→
𝑚
∆t
𝑣→
sun
𝑟→ 𝐹→
∆𝐴
𝑣→
planet
P’](https://image.slidesharecdn.com/gravitationppt-180122143106/85/GRAVITATION-CLASS-11TH-38-320.jpg)
![ ∆𝐴
→
=
1
2
𝑟→ x
𝑃→
𝑚
∆t
∆𝐴
→
∆t
=
1
2
𝑟→ x
𝑃→
𝑚
∆𝐴
→
∆t
=
1
2𝑚
(𝑟→
x 𝑃→
)
∆𝐴
→
∆t
=
𝐿
→
2𝑚
∆𝐴
→
∆t
= constant [L and m are constant]](https://image.slidesharecdn.com/gravitationppt-180122143106/85/GRAVITATION-CLASS-11TH-39-320.jpg)
Uploaded byHIMANSHU .
GRAVITATION CLASS 11TH
This document provides an overview of key concepts in gravitation including: the definition of gravitation; Newton's law of universal gravitation; acceleration due to gravity and how it varies with height and depth; escape velocity; orbital velocity; gravitational potential; time period of satellites; Kepler's laws of planetary motion; and types of satellites. Key points covered include how gravity decreases with height but increases with depth below the Earth's surface, and definitions of geostationary, polar, and binding energy as they relate to satellites orbiting the Earth.
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- 1.
- 2. Definition forGravitation Acceleration Due to Gravity Variation Of “G” With Respect to Height And Depth Escape Velocity Orbital Velocity Gravitational Potential Time period of a Satellite Height of Satellite Binding Energy Various Types of Satellite Kepler’s Law of Planetary motion
- 3. • Gravitation isan Natural Phenomenon by physical bodies attract each other. • Newton focused on his attention on the motion of the moon about the earth. • He declared that the laws of nature that operating between earth and the apple is same as the laws operating between earth and moon. • He calculated the acceleration due to gravitation (on earth) to be 9.8m/s2 . • Newton predicted that a ∝ 1 𝑟2
- 4. • Also, • F∝ 𝑚 𝑟2 • By , The third law of motion, the force on a body due to the earth must be equal to the force on the earth due to the body. Therefore, this force should also be proportional to the mass of the earth. Thus , the Force between the earth and a body is • F ∝ 𝑀𝑚 𝑟2 • F= 𝐺𝑀𝑚 𝑟2 • G is called the universal constant of gravitation and its value is found to be 6.67x10-11 N-m2 /kg2 .
- 5. The rotationof the earth around the sun or that of the moon around earth is explained on the basis of this law. The tide are formed in ocean due to the gravitational force of attraction between earth and the moon. The value of g can be used to predict the orbits and time period of an satellite. Attraction Force between two bodies
- 6. The Gravitationforce between two masses is independent of the intervening medium. The mutual gravitational forces between two bodies are equal and opposite i.e. Gravitational Forces obey Newton’s third law of motion. The gravitational force is an conservative force. The law of gravitation holds only for point masses. The gravitational force between two point masses is a central force. Its magnitude depends only on r and has no angular dependence . The Gravitational force between two bodies is independent of the presence of other bodies.
- 7. Consider anElevator freely falling from North pole to South pole through the center of earth. Describe its motion?? For answer Press Enter ..
- 8. Suppose amass ‘m’ is situated outside the earth at a distance ‘r’ from it centre. The gravitational force on the mass is F = 𝑚𝑣 𝑟2 F=mg. mg= 𝐺𝑀𝑚 𝑟2 g= 𝐺𝑀 𝑟2 By the above equation we can say that Acceleration due to gravity is independent of mass.
- 9. Determine themass of earth using the knowledge of G. G=6.6x1011 and g=9.8 m/s2 and r = 6.36x106 . For answer press enter… As we know g= 𝐺𝑀 𝑟2 . m= 𝑔𝑟2 𝐺 . m= 9.8 ∗(6.96∗106 )2 𝐺 m=5.97x1024 Do you know ?? Cavendish was the first person the calculate the mass of earth because he was the first person to calculate value of G.
- 10. Consider aBody placed at the point ‘p’ above the earth surface at a distance of ‘h’. The g will be . g = 𝐺𝑀 𝑟2 at earth surface. …(i). gh = 𝐺𝑀 𝑟+ℎ 2 at point m ….(ii). Dividing (ii) by (i). We get 𝑔ℎ ℎ = 𝐺𝑀 𝑟+ℎ 2 𝐺𝑀 𝑟 2 𝑔ℎ ℎ = 𝑟2 𝑟+ℎ 2…..(iii) 𝑔ℎ ℎ = 𝑟2 𝑟2 (1+ ℎ 𝑟 )2 𝑔ℎ ℎ = (1 + ℎ 𝑟 )-2
- 11. 𝑔ℎ 𝑔 = 1−2ℎ 𝑟 …(iv) Both theeq. (iii and iv) show that the ‘g’ decreases as the height increases. Eq. iii must be used in the numerical when ‘h’ is comparable to ‘r’. Eq. iv must be used in the numerical when h<<r. Variation of g with respect to the altitude.
- 12. Consider aBody placed at the point ‘d’ below the earth surface at a distance of ‘d’. The g will be . g = 𝐺𝑀 𝑟 2 M = Volume x Density = 4𝜋𝑟3 𝜌. g = 𝐺 4 3 𝜋𝑟3 𝜌 𝑟2 g = 𝐺 4 3 𝜋𝑟𝜌. At earth surface ...(i) gd= 𝐺 4 3 𝜋(𝑟 − 𝑑)𝜌. At the point d …(ii) Dividing (ii) by (i). gd/g = G4/3π(r-d)ρ / G4/3πrρ gd/g = g(1-d/r). Clearly the “g” increases with increase in depth.
- 13. The Variationof g with respect to Depth and Height (graph). The Variation of g with respect to height will never be zero. It will increase after certain height. The Variation of g with respect to depth will be zero when the body is placed at the center of earth.
- 14. Earth isnot perfectly sphere. It is flattened at the poles and bulges out at equator. g = 𝐺𝑚 𝑅2 g ∝ 1 𝑅2 Type equation here.. The radius at Equator is greater than the radius at pole by 21 km. Therefore Ge > Gp. The variation of g between the poles and the equator is about 0.5%.
- 15. Gravitational field Gravitationalfield . Two bodies attract each other by the gravitational force even if they are not in direct contact. This interaction is called action at a distance. It can best explained in terms of concept of field. According to the field concept. Every mass modifies the space around it . This modified space is called gravitational field When any other mass is placed in this field , it feels a gravitational force of attraction due to its interaction with the gravitational field The space surrounding a material body within which its gravitational force of attraction can be experiences a force of attraction towards the centre of earth
- 16. Gravitational Potential Energy Amount of work done in bringing a body from infinity to the given point in the gravitational field of the other. Expression for Gravitational potential energy. F = 𝐺𝑀𝑚 𝑥2 Work down in bringing the body to point B from Point A. W = Fdx W = 𝐺𝑀𝑚 𝑥2 dx 𝑤 = ∞ 𝑟 𝐺𝑀𝑚 𝑥2 dx W = GMm ∞ 𝑟 x−2 W = -GMm ∞ 𝑟 x−1 A P R O r x B dx
- 17. W =-GMm [ 1 𝑅 - 1 ∞ ] U = - 𝐺𝑀𝑚 𝑟 Gravitational Potential Amount of work done in bringing a body of unit mass from infinity to the given point Gravitational Potential = - 𝐺𝑀 𝑟
- 18. If wethrow a ball into air , it rises to a certain height and falls back. If we throw it with a greater velocity , it will rise higher before falling down. If we throw with sufficient velocity , it will never come back . i.e. It will escape from the gravitational pull of the earth. The minimum velocity required to do so is called escape velocity. Consider the earth to be a sphere of mass M and radius R with centre O.
- 20. Using totalenergy concept we can derive the equation for escape velocity. At any point the Total energy must be zero. We know that Kinetic energy = ½ 𝑚𝑣𝑒2 Potential energy = −𝐺𝑀𝑚 𝑅 K.E+P.E = 0 ½ 𝑚𝑣𝑒2 + −𝐺𝑀𝑚 𝑅 =0 ve 2 = 2𝐺𝑀 𝑅 Ve = 2𝐺𝑀 𝑅 Multiply and divide by R Ve = (2gR)
- 21. Can youtell why moon has no atmosphere ?? For answer press enter Due to the small value of g. The escape velocity in the moon is 2.38 km/s . The air molecules have thermal velocity is greater than the escape velocity and therefore air molecules escape.
- 22.
- 23. • Satellite isan body which continuously revolves on it own around and a much larger body in a stable orbit. • Natural satellites : A satellite created by nature is called natural satellite . example : moon. • Artificial satellite : A man made satellite is called an artificial satellite. Example Chandrayaan . • World’s Frist satellite was SPUTNIK-1.
- 24. • Principle forlaunching a satellite : Consider a high tower with its top projecting outside the earth’s atmosphere. • Lets throw a body horizontally from the top of the tower with different velocities. • As we increase the velocity of horizontal projection , the body will hit the ground at point farther and farther from the foot of the tower. • At certain velocity the body will not hit the ground , but always be in a state of free fall under the influence of the gravity. • Then the body will follow a stable circular orbit . And that body is called satellite.
- 25. Click on thevideo
- 26. • Orbital velocityis the velocity required to put the satellite into its orbit around earth
- 27. Force ofgravity on satellite F = 𝐺𝑀𝑚 (𝑅+ℎ)2 Centripetal Force required by the satellite to keep its orbit F = 𝑚𝑣 𝑜 2 (𝑅+ℎ)2 In equilibrium , the centripetal force is just provided by the gravitational pull of the earth. 𝑚𝑣 𝑜 2 (𝑅+ℎ) = 𝐺𝑀𝑚 (𝑅+ℎ)2 Vo 2 = 𝐺𝑀 (𝑅+ℎ) Vo = 𝐺𝑀 (𝑅+ℎ) simplified eq is 𝐺𝑀 𝑅 R R+h h Vo
- 28. • It isthe time takes by a satellite to complete one revolution around the earth. It is given by • T = circumference of the orbit / orbital velocity • T = 2𝜋(𝑅+ℎ) 𝐺𝑀 𝑅+ℎ • T=2𝜋 𝑅+ℎ 3 𝐺𝑚 • If the earth is a sphere the density = 𝜌 then mass would be = volume * density = 4 3 𝜋𝑅3 𝜌 • T= 2𝜋 𝑅+ℎ 3 𝐺 4 3 𝜋𝑅3 𝜌 = 3𝜋 𝑅+ℎ 3 𝐺𝜌𝑅3
- 29. • T=2𝜋 𝑅+ℎ 3 𝐺𝑚 (Gm= gR2) • T2=4𝜋2 𝑅+ℎ 3 𝑔𝑅2 • (R+h)3 = 𝑇2 𝑅2 𝑔 4𝜋2 • R+h = 3 𝑇2 𝑅2 𝑔 4𝜋2 • H = 3 𝑇2 𝑅2 𝑔 4𝜋2 - R
- 30. • A satellitewhich revolves around the earth in tis equatorial plane with the same angular speed and in the same direction as the earth rotates about its own axis is called a geostationary or synchronous satellite.
- 31. • It shouldrevolve in an orbit concentric and coplanar with the equatorial plane of the earth. • Its sense of rotation should be same as that of the earth , i.e From west to east. • Its period of revolution around the earth should be exactly same as that of the earth about its own axis , i.e 24 hours • It should revolve at a height of exactly 35930 km.
- 32. • In communicatingradio,T.V and telephone signals across the world. • In studying the upper regions of the atmosphere. • In Forecasting weather. • In studying meteorites. • In studying solar radiation and cosmic rays. • And used in GPS (Global positioning System).
- 33. • A satellitethat revolves in a planar orbit is called a polar satellite. Eg IERS (Indian earth resources satellites) • Uses of Polar satellite – Polar satellites are used in weather and environment monitoring. – Spying – Study topography of other celestial bodies
- 34. • Consider asatellite of mass m moving around the earth with velocity in an orbit of a radius r. • Because of the gravitation pull of the earth, the satellite has a potential energy which is given by • U = - 𝐺𝑀𝑚 𝑟 • Kinetic energy = 1 2 m 𝐺𝑀 𝑅 . • Total Energy = - 𝐺𝑀𝑚 𝑟 + 1 2 m 𝐺𝑀 𝑅 • E = - 𝐺𝑀𝑚 2𝑟 • The negative sign shows that the satellite is bound to the earth.
- 35. • The energyrequired by a satellite to leave its orbit around the earth and escape to infinity is called binding energy. • Binding energy = + 𝐺𝑀𝑚 2𝑟 • Because the total energy of the satellite is - 𝐺𝑀𝑚 2𝑟 in order to escape into infinity , it must be supplied extra energy so that its energy E becomes zero.
- 36. • Law oforbits (first law) : Each planet revolves around the sun in an elliptical orbit with the sun situated at the one of the two foci.
- 37. • Law ofareas (second law) : The radius vector drawn from the sun to a planet sweeps out equal areas in equal intervals of time i.e the areal velocity ( area covered per unit time) of a planet around the sun is constant.
- 38. Proof : 𝜏 = 𝑟→ x 𝐹→ = 0 [ opp. Faced] 𝜏 = 𝑑𝐿→ 𝑑𝑡 = 0. Or 𝐿→ = constant . ∆𝐴 → = 𝐴𝑟𝑒𝑎 𝑜𝑓 𝑡ℎ𝑒 𝑡𝑟𝑖𝑎𝑛𝑔𝑙𝑒 ∆𝐴 → = 1 2 𝑟→ x 𝑃→ P (∆𝐴 → = Area) 𝑃→ P = ∆ 𝑟→ = 𝑣→ ∆t = 𝑃→ 𝑚 ∆t 𝑣→ sun 𝑟→ 𝐹→ ∆𝐴 𝑣→ planet P’
- 39. ∆𝐴 → = 1 2 𝑟→ x 𝑃→ 𝑚 ∆t ∆𝐴 → ∆t = 1 2 𝑟→x 𝑃→ 𝑚 ∆𝐴 → ∆t = 1 2𝑚 (𝑟→ x 𝑃→ ) ∆𝐴 → ∆t = 𝐿 → 2𝑚 ∆𝐴 → ∆t = constant [L and m are constant]
- 40. • Law ofperiods (third law) : The square of the period of revolution of a planet around the sun is proportional to the cube of the semi major axis of the its elliptical orbit. • Proof : Suppose a planet of mass m moves around the sun in a circular orbit of radius r with orbital speed v. Let M be the mass of the sun. The force of gravitation between the sun and the planet provides the necessary centripetal force. • 𝑚𝑣2 𝑟 = 𝐺𝑀𝑚 𝑟2 or v2 = 𝐺𝑀 𝑟 . • v = 𝑐𝑖𝑟𝑐𝑢𝑚𝑓𝑒𝑟𝑒𝑛𝑐𝑒 𝑃𝑒𝑟𝑖𝑜𝑑 𝑜𝑓 𝑟𝑒𝑣𝑜𝑙𝑢𝑡𝑖𝑜𝑛 = 2𝜋𝑟 𝑡 • 4𝜋2 𝑟2 𝑡2 = 𝐺𝑀 𝑟
- 41. 𝑡2= 4𝜋2 𝐺𝑀 x r3= constant r3 𝑡2 ∝ r3 . For better understanding please click the here.
- 42.