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10.2 fields at work 2015

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10.2 fields at work 2015

  1. 1. 10.2 FIELDS AT WORK
  2. 2. GRAVITATIONAL POTENTIAL AND STRENGTH THE GRAVITATIONAL FIELD STRENGTH IS A VECTOR QUANTITY WHOSE DIRECTION IS GIVEN BY THE DIRECTION OF THE FORCE A MASS WOULD EXPERIENCE IF PLACED AT THE POINT OF INTEREST G = G M R2 COMPARE THIS TO GRAVITATIONAL POTENTIAL V = - G M R AND WE GET THE RELATIONSHIP THAT G = -V R
  3. 3. ELECTRIC AND GRAVITATIONAL FIELDS. (A) INVERSE SQUARE LAW OF FORCE COULOMB'S LAW IS SIMILAR IN FORM TO NEWTON'S LAW OF UNIVERSAL GRAVITATION. BOTH ARE INVERSE SQUARE LAWS WITH CONSTANTS OF 1/(4ΠΕ) IN THE ELECTRIC CASE CORRESPONDING TO THE GRAVITATIONAL CONSTANT G. THE MAIN DIFFERENCE IS THAT WHILST ELECTRIC FORCES CAN BE ATTRACTIVE OR REPULSIVE, GRAVITATIONAL FORCES ARE ALWAYS ATTRACTIVE. TWO TYPES OF ELECTRIC CHARGE ARE KNOWN BUT THERE IS ONLY ONE TYPE OF GRAVITATIONAL MASS. BY COMPARISON WITH ELECTRIC FORCES, GRAVITATIONAL FORCES ARE EXTREMELY WEAK.
  4. 4. (B) FIELD STRENGTH THE FIELD STRENGTH AT A POINT IN A GRAVITATIONAL FIELD IS DEFINED AS THE FORCE ACTING PER UNIT MASS PLACED AT THE POINT. THUS IF A MASS M IN KILOGRAMS EXPERIENCES A FORCE F IN NEWTONS AT A CERTAIN POINT IN THE EARTH'S FIELD, THE STRENGTH OF THE FIELD AT THAT POINT WILL BE F/M IN NEWTONS PER KILOGRAM.
  5. 5. THIS IS ALSO THE ACCELERATION A, THE MASS WOULD HAVE IN METRES PER SECOND SQUARED IF IT FELL FREELY UNDER GRAVITY AT THIS POINT (SINCE F = MA). THE GRAVITATIONAL FIELD STRENGTH AND THE ACCELERATION DUE TO GRAVITY AT A POINT THUS HAVE THE SAME VALUE (I.E. F/M) AND THE SAME SYMBOL, G, IS USED FOR BOTH. AT THE EARTH'S SURFACE G = 9.8 N KG-' = 9.8 M S-2 (VERTICALLY
  6. 6. (C) FIELD LINES AND EQUIPOTENTIALS THESE CAN ALSO BE DRAWN TO REPRESENT GRAVITATIONAL FIELDS BUT SUCH FIELDS ARE SO WEAK, EVEN NEAR MASSIVE BODIES, THAT THERE IS NO METHOD OF PLOTTING FIELD LINES SIMILAR TO THOSE USED FOR ELECTRIC (AND MAGNETIC) FIELDS. FIELD LINES FOR THE EARTH ARE DIRECTED TOWARDS ITS CENTRE AND THE FIELD IS SPHERICALLY SYMMETRICAL. OVER A SMALL PART OF THE EARTH'S SURFACE THE FIELD CAN BE CONSIDERED UNIFORM, THE LINES BEING VERTICAL, PARALLEL AND EVENLY SPACED.
  7. 7. (D) POTENTIAL AND P.D. ELECTRIC POTENTIALS AND PDS ARE MEASURED IN JOULES PER COULOMB (J C-1) OR VOLTS; GRAVITATIONAL POTENTIALS AND PDS ARE MEASURED IN JOULES PER KILOGRAM (J KG-1). AS A MASS MOVES AWAY FROM THE EARTH THE POTENTIAL ENERGY OF THE EARTH-MASS SYSTEM INCREASES, TRANSFER OF ENERGY FROM SOME OTHER SOURCE BEING NECESSARY.
  8. 8. IF INFINITY IS TAKEN AS THE ZERO OF GRAVITATIONAL POTENTIAL (I.E. A POINT WELL OUT IN SPACE WHERE NO MORE ENERGY IS NEEDED FOR THE MASS TO MOVE FURTHER AWAY FROM THE EARTH) THEN THE POTENTIAL ENERGY OF THE SYSTEM WILL HAVE A NEGATIVE VALUE EXCEPT WHEN THE MASS IS AT INFINITY. AT EVERY POINT IN THE EARTH'S FIELD THE POTENTIAL IS THEREFORE NEGATIVE (SEE EXPRESSION BELOW), A FACT WHICH IS CHARACTERISTIC
  9. 9. ESCAPE SPEED THE ESCAPE SPEED IS THE SPEED REQUIRED FOR A PROJECTILE TO LEAVE THE EARTH´S GRAVITATIONAL ATTRACTION. I.E. TO GET TO INFINITY!
  10. 10. IF THE POTENTIAL AT THE EARTH´S SURFACE IS V = - G ME RE THEN THE EP CHANGE TO GET TO INFINITY IS G ME X M RE WHERE M IS THE MASS OF THE PROJECTILE
  11. 11. FOR THIS AMOUNT OF ENERGY TO BE GAINED THE PROJECTILE MUST HAVE HAD AN EQUAL AMOUNT OF EK THEREFORE ½MV2 = G ME X M RE V = ( 2GME RE ) BUT USING THE FACT THAT G = G ME RE 2 THEN V = (2GRE )
  12. 12. DERIVING THE THIRD LAW SUPPOSE A PLANET OF MASS M MOVES WITH SPEED V IN A CIRCLE OF RADIUS R ROUND THE SUN OF MASS M, THE GRAVITATIONAL ATTRACTION OF THE SUN FOR THE PLANET IS = G MM R2 FROM NEWTON’S LAW OF UNIVERSAL GRAVITATION IF THIS IS THE CENTRIPETAL FORCE KEEPING THE PLANET IN ORBIT THEN G MM = MV2 (FROM CENTRIPETAL EQUATION) R2 R GM = V2 R
  13. 13. IF T IS THE TIME FOR THE PLANET TO MAKE ONE ORBIT V = 2 R V2 = 222 R2 T T2 GM = 4 2 R 2 R T2 GM = 4 2 R 3 T2 R 3 = GM T2 4 2 R 3 = A CONSTANT T2
  14. 14. ENERGY OF ORBITING SATELLITES POTENTIAL ENERGY, EP A SATELLITE OF MASS M ORBITING THE EARTH AT A DISTANCE R FROM ITS CENTRE HAS GRAVITATIONAL POTENTIAL, V = - G ME R THEREFORE THE GRAVITATIONAL POTENTIAL ENERGY, EP = - G MEM R
  15. 15. ENERGY OF ORBITING SATELLITE KINETIC ENERGY, EK BY THE LAW OF UNIVERSAL GRAVITATION AND NEWTON’S SECOND LAW G MEM = MV2 R2 R THEREFORE THE KINETIC ENERGY, EK = ½MV2 = G MEM 2R
  16. 16. ENERGY OF ORBITING SATELLITE TOTAL ENERGY, EP+ EK TOTAL ENERGY = - G MEM + G MEM R 2R TOTAL ENERGY = - G MEM 2R TOTAL ENERGY IS CONSTANT FOR A CIRCULAR ORBIT.
  17. 17. GRAPHS OF ENERGY AGAINST RADIUS • POTENTIAL ENERGY Ep 1/r Ep r i.e. Ep  -1/r or Ep = -k/r where k is the constant of proportionality = GMem
  18. 18. GRAPHS OF ENERGY AGAINST RADIUS • KINETIC ENERGY Ek 1/r Ek r i.e. Ek  1/r or Ek = k/r where k is the constant of proportionality = GMem 2
  19. 19. GRAPHS OF ENERGY AGAINST RADIUS • TOTAL ENERGY Etotal 1/r Etotal r i.e. Etotal  -1/r or Etotal = -k/r where k is the constant of proportionality = GMem 2
  20. 20. 1. FIND THE GRAVITATIONAL POTENTIAL 1000KM ABOVE THE EARTH’S SURFACE? 2. A SATELLITE OF MASS 50KG MOVES FROM A POINT WHERE THE POTENTIAL IS –20 MJKG-1 TO ANOTHER POINT WHERE THE POTENTIAL IS –60 MJKG-1 • WHAT IS THE CHANGE IN POTENTIAL • WHAT IS THE SPEED OF THE SATELLITE 3. A 2000KG SPACECRAFT IN ORBIT AT R ABOVE THE EARTH OF RADIUS R. THE POTENTIAL AT THE EARTH'S SURFACE IS –60 MJ KG-1. WHAT IS THE CHANGE IN POTENTIAL ENERGY IF THE SPACECRAFT RETURNS TO EARTH

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