Calculation of the Gravitational Constant. Abstract. + α R 2 1. T 3 cos 12 β

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1 Calculation of the Gravitational Constant Abstract G 4 M ( π 5 α 2 ω + α R 2 tan θ R 2 2 ω 20T 5 cos 20 β π 3 α 2 ω 8 ω 2G4 T 3 cos 2 β G is the Gravitational constant and M is the mass of Earth. ω Angular speed of revolution of Moon rad/s ω Angular speed of rotation of Earth rad /s R 2 Square of the radius of Earth m 2 R Radius of Moon m T Time during 2 synodical period of Moon seconds α The ratio of mass of Moon to the mass of Earth α β θ Angle between the plane of the orbit of Moon and the equatorial plane of Earth Hence we get and G M

2 Calculation of the Gravitational Constant. Angular Speed of Revolution of Moon : The time period of revolution of Moon is T days One day 23 hours 56 minutes seconds hours So, the angular speed of revolution of Moon is w 2π T rad/s 2. Angular Speed of Rotation of Earth : - The time period of rotation of Earth is T 23 hours 56 minutes seconds hours So, the angular speed of rotation of Earth is w 2π T rad/s 3. Mass of Moon :- The mass of Earth is and the mass of Moon is M kg M kg The ratio of mass of Earth to the mass of Moon is M M M

3 We determine the mass of Moon by using the formula M w 2 R cos θ w 2 2 a where M is the ratio of mass of Earth to the mass of Moon, w is the angular speed of rotation of Earth, w 2 is the angular speed of revolution of Moon, a is the mean distance of Moon from Earth, R is the equatorial radius of Earth and θ is the angle between the plane of the orbit of Moon and the equatorial plane of Earth. cosθ Mw 2 2 a w 2 R...() 4. Radius of Earth :- The equatorial and polar radii of Earth are respectively given by a m and So, the mean radius of Earth is b m R 2a + b m Let α be the eccentric angle of the mean radius R of Earth R 2 a 2 cos 2 α + b 2 sin 2 α cos 2 α R2 b2 a 2 b cosα α o is the eccentric angle of the mean radius of Earth. The angle between the planes of the orbits of Moon and Sun o o o α (s a y) 3

4 The angle of inclination of Moon is α 2 o o α (α + α 2 ) o θ(s a y) The radius R of Earth for its ecentric angle θ is given by 5. Radius of Moon : - R 2 a 2 cos 2 θ + b 2 sin 2 θ m 2 Let a be the mean distance of Moon from Earth. The radius of Moon at its mean distance from Earth is Hence, the radius of Moon is θ o R a θ π 80 o 6. Mean Distance of Moon from Earth : - The length of the mean sidereal day is The length of the mean solar day is t 23 hours 56 minutes seconds hours t 2 24 hours 3 minutes seconds hours The length of the nodical month ( node to node ) is T days The length of the sidereal month (fixed star to fixed star) is T days 4

5 The geometric mean of t t 2 and T T 2 is α t t 2 T T The distance d of Moon from Earth is given by d 3 GM w 2 where M is the mass of Earth, w is the angular speed of revolution of Moon and G being the Gravitational constant. d (GM) 3 Hence, the mean distance of Moon from Earth is w Gravitational Constant:- Now, a d α (GM) 3 αw (2) w 2 a α g 2 and w α g 2 where w is the angular speed of revolution of Moon, a is its mean distance from Earth and g is the acceleration due to gravity of both Earth and Moon. where G is the Gravitational constant. We write, G w3 a g 2 g g 0 g tan θ 2 where g is the acceleration due to gravity of both Earth and Moon, g 0 is the acceleration due to gravity of Earth, g is the acceleration due to gravity of Moon and θ is the angle between the plane of the orbit of Moon and the equational plane of Earth. 5

6 The mass of Earth is and the mass of Moon is M kg M kg The ratio of mass of Moon to the mass of Earth is We have α M M g g 0 + g tan θ 2 GM R 2 + GM tan θ 2 R 2 where R is the radius of Earth and R g GM R 2 is the radius of Moon + Gα M R 2 tan θ 2 Hence, G w3 a g 2 ( GM R 2 w 3 a + Gα M tan θ R 2 2 w 3 (GM) 3 ( ) αw 2 2 (using equation (2) ) 3 G 2 M 2 + α R 2 tan θ R 2 2 G 8 w ( ) αm (3) 3 + α R 2 tan θ R 2 2 The unit of w3 a in the R.H.S. of the equation g2 G w3 a g 2 is s 3.m (m/s 2 s m 6

7 But, the unit of the Gravitational constant G is N.m 2 kg 2 kg.m s 2. m2 kg 2 m3 s 2.kg So, in order to transfer s/m We write, to m3 s 2.kg ; we should have to multiply s/m (w a cos β) 4 T πm with m4 s 3.kg where M is the mass of Earth, w is its angular speed of rotation, a is the mean distance of Moon from Earth, T being the time during 2 synodical period of Moon and β is an angle of derivation. Thus, the unit of the L.H.S. of the above equation becomes m 4 s 3.kg. Hence, multiplying it with a w3 of the equation g2 we get the desired unit of G G w3 a g 2 without the change of its numerical value. Henceforth, there should not be any confusion about the dimension of G. As a result of it we get M (w a cos β) 4 T π w4 (GM)4 3 T cos 4 β (using equation (2)) πα 4 w 8 3 M 3 w 4 G 4 3 T cos 4 β πα 4 w 8 3 or, M 3 πα 4 w 8 3 w 4 G 4 3 T cos 4 β (4) M 5 π 5 α 20 w w 20G 20 3 T 5 cos 20 β (5) 7

8 Using equation (5) in equation (3) we have G 8 w 7 3 w 20 3 T 5 cos 20 β 3 ( απ 5 α 20 w α R 2 tan θ R 2 2 G 4 G 20 w 20 G 20 3 T 5 cos 20 β ( π 5 α 2 w + α R 2 tan θ R 2 2 ( π 5 α 2 w + α R 2 tan θ R 2 2 w 20T 5 cos 20 β (6) Let us derive the angle β. Recall that we have the radius R R m 2 of Earth given by and the eccentric angle of R is β o The angle of inclination of Moon is β 2 o o The angle between the geographic and magnetic meridians of Earth is β 3 7 o The angle between the planes of the orbits of Moon and Sun is β o The synodical period of Moon is β β 4 2 cos(β 3 + β 2 cos β ) o T days One day 23 hours 56 minutes seconds hours T 2 T seconds 8

9 The ratio of mass of Earth to the mass of Moon is M The equatorial radius of Earth is R m w rad/s w rad/s α α R m 2 From equation () we have cosθ M0 w 2 a w 2R 0 M0 w 2 G 3 M 3 w 2R 0αw 2 3 M0 w 4 3 G 3 M 3 w 2 R 0α (using equation(2)) where M0 w 4/3 G 3 πα 4 w 8 3 w 2R 0αw 4G 4 3 T cos 4 β M0 w 4 πα 3 Gw 6R 0T cos 4 β φ G (s a y) M0 w 4 πα 3 φ w 6R 0T cos 4 β (using equation (4)) From equation (6) we have G 4 π5 α 2 w ( w 20T 5 cos 20 β R 2 + α R 2 tan θ 2 φ 2 ( R 2 + α R 2 tan θ 2 (s a y) 9

10 where φ 2 π5 α 2 w w 20 T 5 cos 20 β The radius of Moon at its mean distance a from Earth is ψ o R a ψ α R 2 π 80 o G 3 M 3 πψ αw (using equation (2)) G 3 πα 4 w 8 3 πψ αw 2 3 w 4G 4 3 T cos 4 β 80 π 2 α 3 w 2 ψ G 80 w 4 T cos4 β G2 ( α w 8 T 2 cos 8 β π 4 α 6 w 4 ψ 2 φ 3 G 2 (s a y) (using equation (4)) where φ 3 (.8)2 0 4 α w 8 T 2 cos 8 β π 4 α 6 w 4 ψ

11 Hence, or, or, On iteration it yields the ( G 4 φ 2 R 2 + φ 3G 2 tan θ 2 ( cosθ φ 2 R 2 + φ 3G 2 + cosθ ( φ 2 R 2 + φ 3G 2 φ /G + φ /G ( φ 2 R 2 + φ 3G 2 G φ G + φ G 2 ( ) φ 2 R 2 + φ 3G 2 G φ G + φ G 2 φ2 R 2 φ 3G 2 G φ G + φ φ2 R 2 G 2 φ G φ 3 G + φ L.H.S ( cosθ φ ) G and the for the value of R.H.S G Hence, the Gravitational constant is The value of the Gravitational constant G G N.m 2 /kg 2 is correct provided for the given ratio of mass of Earth equal to kg to the mass of Moon equal to kg. 8. Mass of Earth :- From equation (4) we have M 3 πα 4 w 8 3 w 4 G 4 3 T cos 4 β M π 3 α 2 w 8 w 2 G4 T 3 cos 2 β

12 is the mass of Earth in kg. 9. Mass of Moon : - The mass of Moon is M M M kg 0. Mean Distance of Moon from Earth :- The mean distance of Moon from Earth is a ( GM α w 2 ) m. Radius of Moon :- The radius of Moon at its mean distance a R a ψ π 80 o from Earth is m 2. Acceleration due to Gravity of Moon :- The acceleration due to gravity of Moon is g GM R m/s 2 3. Value of θ:- We have M0 w 2 a cosθ w 2R θ o 4. Value of g :- We have G w3 a g g 2 w 3 a G m/s 2 2

13 5. Value of g 0 :- We have g 0 GM R m/s 2 6. Value of g tan θ 2 :- g tan θ 2 GM R 2 tan θ m/s 2 7. Conclusion - The Gravitational constant is G s/m or, G N.m 2 /kg 2 References [] College Physics by Weber, Manning & White [2] Clark s Tables [3] Siddhanta Darpana [4] Indian Astronomical Ephemeris, [5] Pears Cyclopaedia 3