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Newton’s Law and Planetary Fields
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Kepler's Third → Kepler's Third Law of planetary motion states that 7 IS the Table 1 Kepler's third law Mercury Equation Sheet Sub Law та Newton's Law of Gravitation F = · Newton used Kepler's Third Law and used the inverse- square law of force to devise Newton's law of gravitation. 9 = F gravitational force m GMm into : 3 g= GMP GM Gravitational Field Strength from Newton's Law X . 3 in Newtons's Law of Gravitation & Planetary Fields G m. m₂ r₂ Magnitude of gravitational field strength a radial field: 6.67x10-¹¹ Nm ²kg (where m <<M) Equation Sheet distance aport (3) Average radius r of orbit/1010 m 6 Time T for one orbit/10's 2/1015 m³s-2 masses of two objects -2 GM 6 0.8 337 game Venus Earth 11 1.95 mass m₂ 350 M distance/m from contre Torsion wire K L/2 for all M gravitational constant big mass / kg 15 3.2 330 planets. Mars 23 5.9 349 Jupiter Cavendish Experiment to find G 78 37.4 340 distance r F+ Saturn point of mass M 143 93.0 338 measuring the angle it twisted rough, the force of attraction between each massive lead ball and the small ball hearest to it was calculated. planet of mass M + mass m₂ ▲ Figure 1 Comparing fields The Variation of Gravitational Field Strength g gs Sub O 2 R 3R 48 R Distance from centre of planet of radius R Deriving g within planet V = р = with Distance From Centre of Planet 9 : F into w = FS g₁ = g surface M = PV M = is Tvr³ TV Caloulating Gravitational Potential p. ³ and variable E P Ep F = GPE = SF ds •The shape of inverse square law Proportion to r². This is be cause GMm r² • Why does GMm ..Gravitational potentials Ep = [-GMmr"] GMM dr !! The potential energy (E₁) is defined as object moves to its position from infinity. (in this case 2) curve beyond surface of planet is curve because decreases 9 → As r decreases, M decreases. • This is because only the mass contributes to g. The remainder of resultant force. . At centre. is 0 as...
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Alternative transcript:
mass 9 R + decrease linearly inside a GM g= ra r ...O are always negative mass g Subbing for M: = 9 = energy gained as g= 9 = G from GM 4TG pr 3 4π Gp in the sphere of radius r r < x < R does not give • p. ³ r homogenous planet? (r< R) V = V = that 9 Equation sheet = contributes to is O. an in inverse (for r <R) GM X mass of GM planet/kg distance/m !! Carefu r not squared Escape Velocities Seen above In moving kinetic energy : Total Ep &w= Ep " = work done in moving from = for = a small of = 1 Integrating to Gravitational Field Strength g gs R G - kg Aw= FS G Mm ra distance &r against gravity (GMm), work done the body is given by W = FS: GMm [===// ] Mm R Gravitational potential at 1 kg mass W = GMm R to given → The escape velocity is the minimum velocity an object must be the planet when projected vertically from the surface. G Mm dr r' GM R = IVI X GM R where 9 = 9₁R² 12 R as &r to infinity: B R R is find gravitational potential at surface move from infinty to surface V surface m = 1 :. At surface: radius of Earth .. V= Plotting g against r g 9₁ = GM = O R 2 R 4R 8 R Distance from centre of planet of radius R -GM surface : Why does it GM R² go R² g&R ² give xr AW m = V the body is of work, it needs at the moment it If AW to move V surface surface potential V surface ? :: 1/2mv ² -DV g= or ov=gAr C to be able to at least this projected = > V² » Vesc Subbing Ⓒ at the expense of the surface = GM = Vesc = Vesc escape velocity SCR = Gmm R = into 2GM R GM to gs 2 GM R [g. R²r + ] do this amount amount of KE Rª ↑ BORS dr gs R² r² escape from 00 x R 98 INkg" Veso: 298 R² R 2gs R 6-37x106m = [ = 368²²- ] - [ -9² R² ] ·9s P R = -98 R =9.81 X 6.37 × 10 6 = -62 MJ kg"
Newton’s Law and Planetary Fields
Newton’s Law and Planetary Fields
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Kepler's Third → Kepler's Third Law of planetary motion states that 7 IS the Table 1 Kepler's third law Mercury Equation Sheet Sub Law та Newton's Law of Gravitation F = · Newton used Kepler's Third Law and used the inverse- square law of force to devise Newton's law of gravitation. 9 = F gravitational force m GMm into : 3 g= GMP GM Gravitational Field Strength from Newton's Law X . 3 in Newtons's Law of Gravitation & Planetary Fields G m. m₂ r₂ Magnitude of gravitational field strength a radial field: 6.67x10-¹¹ Nm ²kg (where m <<M) Equation Sheet distance aport (3) Average radius r of orbit/1010 m 6 Time T for one orbit/10's 2/1015 m³s-2 masses of two objects -2 GM 6 0.8 337 game Venus Earth 11 1.95 mass m₂ 350 M distance/m from contre Torsion wire K L/2 for all M gravitational constant big mass / kg 15 3.2 330 planets. Mars 23 5.9 349 Jupiter Cavendish Experiment to find G 78 37.4 340 distance r F+ Saturn point of mass M 143 93.0 338 measuring the angle it twisted rough, the force of attraction between each massive lead ball and the small ball hearest to it was calculated. planet of mass M + mass m₂ ▲ Figure 1 Comparing fields The Variation of Gravitational Field Strength g gs Sub O 2 R 3R 48 R Distance from centre of planet of radius R Deriving g within planet V = р = with Distance From Centre of Planet 9 : F into w = FS g₁ = g surface M = PV M = is Tvr³ TV Caloulating Gravitational Potential p. ³ and variable E P Ep F = GPE = SF ds •The shape of inverse square law Proportion to r². This is be cause GMm r² • Why does GMm ..Gravitational potentials Ep = [-GMmr"] GMM dr !! The potential energy (E₁) is defined as object moves to its position from infinity. (in this case 2) curve beyond surface of planet is curve because decreases 9 → As r decreases, M decreases. • This is because only the mass contributes to g. The remainder of resultant force. . At centre. is 0 as...
Kepler's Third → Kepler's Third Law of planetary motion states that 7 IS the Table 1 Kepler's third law Mercury Equation Sheet Sub Law та Newton's Law of Gravitation F = · Newton used Kepler's Third Law and used the inverse- square law of force to devise Newton's law of gravitation. 9 = F gravitational force m GMm into : 3 g= GMP GM Gravitational Field Strength from Newton's Law X . 3 in Newtons's Law of Gravitation & Planetary Fields G m. m₂ r₂ Magnitude of gravitational field strength a radial field: 6.67x10-¹¹ Nm ²kg (where m <<M) Equation Sheet distance aport (3) Average radius r of orbit/1010 m 6 Time T for one orbit/10's 2/1015 m³s-2 masses of two objects -2 GM 6 0.8 337 game Venus Earth 11 1.95 mass m₂ 350 M distance/m from contre Torsion wire K L/2 for all M gravitational constant big mass / kg 15 3.2 330 planets. Mars 23 5.9 349 Jupiter Cavendish Experiment to find G 78 37.4 340 distance r F+ Saturn point of mass M 143 93.0 338 measuring the angle it twisted rough, the force of attraction between each massive lead ball and the small ball hearest to it was calculated. planet of mass M + mass m₂ ▲ Figure 1 Comparing fields The Variation of Gravitational Field Strength g gs Sub O 2 R 3R 48 R Distance from centre of planet of radius R Deriving g within planet V = р = with Distance From Centre of Planet 9 : F into w = FS g₁ = g surface M = PV M = is Tvr³ TV Caloulating Gravitational Potential p. ³ and variable E P Ep F = GPE = SF ds •The shape of inverse square law Proportion to r². This is be cause GMm r² • Why does GMm ..Gravitational potentials Ep = [-GMmr"] GMM dr !! The potential energy (E₁) is defined as object moves to its position from infinity. (in this case 2) curve beyond surface of planet is curve because decreases 9 → As r decreases, M decreases. • This is because only the mass contributes to g. The remainder of resultant force. . At centre. is 0 as...
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Knowunity was a featured story by Apple and has consistently topped the app store charts within the education category in Germany, Italy, Poland, Switzerland and United Kingdom. Join Knowunity today and help millions of students around the world.
Still not sure? Look at what your fellow peers are saying...
iOS User
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Stefan S, iOS User
The application is very simple and well designed. So far I have found what I was looking for :D
SuSSan, iOS User
Love this App ❤️, I use it basically all the time whenever I'm studying
Alternative transcript:
mass 9 R + decrease linearly inside a GM g= ra r ...O are always negative mass g Subbing for M: = 9 = energy gained as g= 9 = G from GM 4TG pr 3 4π Gp in the sphere of radius r r < x < R does not give • p. ³ r homogenous planet? (r< R) V = V = that 9 Equation sheet = contributes to is O. an in inverse (for r <R) GM X mass of GM planet/kg distance/m !! Carefu r not squared Escape Velocities Seen above In moving kinetic energy : Total Ep &w= Ep " = work done in moving from = for = a small of = 1 Integrating to Gravitational Field Strength g gs R G - kg Aw= FS G Mm ra distance &r against gravity (GMm), work done the body is given by W = FS: GMm [===// ] Mm R Gravitational potential at 1 kg mass W = GMm R to given → The escape velocity is the minimum velocity an object must be the planet when projected vertically from the surface. G Mm dr r' GM R = IVI X GM R where 9 = 9₁R² 12 R as &r to infinity: B R R is find gravitational potential at surface move from infinty to surface V surface m = 1 :. At surface: radius of Earth .. V= Plotting g against r g 9₁ = GM = O R 2 R 4R 8 R Distance from centre of planet of radius R -GM surface : Why does it GM R² go R² g&R ² give xr AW m = V the body is of work, it needs at the moment it If AW to move V surface surface potential V surface ? :: 1/2mv ² -DV g= or ov=gAr C to be able to at least this projected = > V² » Vesc Subbing Ⓒ at the expense of the surface = GM = Vesc = Vesc escape velocity SCR = Gmm R = into 2GM R GM to gs 2 GM R [g. R²r + ] do this amount amount of KE Rª ↑ BORS dr gs R² r² escape from 00 x R 98 INkg" Veso: 298 R² R 2gs R 6-37x106m = [ = 368²²- ] - [ -9² R² ] ·9s P R = -98 R =9.81 X 6.37 × 10 6 = -62 MJ kg"