Weightlessness and satellites in orbit. Orbital energies

Transcription

1 Weightlessness and satellites in orbit Orbital energies

2 Review PE = - GMm R v escape = 2GM E R = 2gR E Keppler s law: R3 = GM s T 2 4π 2

3 Orbital Motion Orbital velocity escape velocity In orbital motion an object is still under the influence of Earth s gravitational field, which causes it to follow a parabolic path F c = F g mv2 R E = GM Em R E 2 v 2 = GM EmR E R E 2 v = GM R E = g R E

4 What is weightlessness? Consider Dobson inside an elevator which is not moving If he drops a ball, it will accelerate downward at 10 ms -2 as expected. If the elevator is accelerating upward at 2 ms -2, what does Dobson observe the dropped ball s acceleration to be? SOLUTION: Since the elevator is accelerating upward at 2 ms -2 to meet the ball, and the ball is accelerating downward at 10 ms -2, Dobson observes an acceleration of 12 ms -2. If the elevator is accelerating at 2, he observes a = 8 ms -2.

5 Considder an elevator moving down at 10ms 1 If the elevator is accelerating downward at 10 ms -2, what does Dobson observe the dropped ball s acceleration to be? SOLUTION: He observes the acceleration of the ball to be zero! He thinks that the ball is weightless! The ball is NOT weightless, obviously. It is merely accelerating at the same rate as Dobson! This is what we mean by weightlessness in an orbiting spacecraft

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7 Weightlessness Astronaut and spaceship have the same acceleration Acceleration is towards the center of the planet Both fall at the same rate There is no reaction force between astronauts and the spaceship

8 Question We have all seen astronauts experiencing weightlessness. Explain why it only appears that they are weightless. SOLUTION: The astronaut, the spacecraft, and the tomatoes, are all accelerating at a c = g. They all fall together and appear to be weightless. International Space Station

9 Only in deep space which is defined to be far, far away from all masses will a mass be truly weightless. In deep space, the r in F G = GMm is so large R 2 for every m that F G, the force of gravity, is for all intents and purposes, zero.

10 Geostationary Orbit Above plane of equator Satellites in Orbit Does not move when viewed from Earth Receiving dish on Earth does not need to track transmitting antenna on satellite Communication uses

11 Polar Orbits Close to Earth s surface (100km above it) Orbit is over the poles Over 24 hours satellite will view every point on Earth s surface

12 Geosynchronous satellite Orbital period is 24 h Stays in the same area in the sky but wanders a figure 8 shape

13 Example A geostationary satellite has an orbital time of 24h. A) Calculate the distance of the satellite from the surface of the Earth B) Find the gravitational field strength at orbital radius

14 Example A satellite wishes to orbit Earth at a height of 100km above the surface of the earth. Find the speed and acceleration of the satellite. Solution v = GM R v = 6.67x10 11 x5.98x x x10 3 = 7.85 x 10 3 ms 1 a = GM R 2 or a = v2 R a = 9.53ms 2

15 Satellite energy Two speeds: v escape and v orbit ~ 11.2kms 2 ~ 7.8 kms 2 Orbital shape depends on speed: v orbit at 7.8 circular 7.8 < v < 11.2 elliptical v escape at 11.2 parabola v > v escape hyperbola

16 Energies of satellites An orbiting satellite has both kinetic energy and potential energy. The gravitational potential energy of an object of mass m in the gravitational field of Earth is PE = GMm, where M is the mass of the earth. The kinetic energy of an object of mass m moving at speed v is KE = 1 2 mv2. Thus the total mechanical energy of an orbiting satellite of mass m is TE = KE + PE TE = 1 2 mv2 - GMm R R total energy of an orbiting satellite

17 Show that KE of an orbiting satellite having mass m at a distance R from the center of Earth (mass M) is KE = GMm 2R SOLUTION: In circular orbit F c = F g Example mv 2 R = GMm R 2 mv 2 = GMm R 1 2 mv2 = GMm 2R KE = GMm 2R = 1 2 GMm R = 1 2 PE kinetic energy of an orbiting satellite

18 Three expressions for KE Since GM = g 0 R E 2 KE = GMm 2R In summary KE = 1 2 = g 0 R E 2 m 2R PE = GMm 2R = g 0 R E 2 m 2R

19 Example Show that the total energy of an orbiting satellite at a distance r from the center of Earth is TE = GMm 2R SOLUTION: From TE = KE + PE and the expressions for KE and PE we have TE = KE + PE TE = GMm GMm TE TE = 2R R GMm 2GMm 2R = GMm 2R PE= GMm GMm GMm, KE =, TE = R 2R 2R total energy of an orbiting satellite The IBO expects you to derive these relationships.

20 Graphical representation of KE total energy of an orbiting satellite Graph the kinetic energy vs. the radius of orbit for a satellite of mass m about a planet of mass M and radius R. SOLUTION: Use KE = GMm 2R It has a maximum value of KE = GMm 2R. Note that KE decreases with increasing radius. E K GMm 2R R 2R 3R 4R 5R r

21 Graphical representation of PE Graph the PE vs. the radius of orbit for a satellite of mass m about a planet of mass M and radius R. SOLUTION: Use PE = GMm. Note that PE increases with increasing radius. It becomes less negative. R total energy of an orbiting satellite R 2R 3R 4R 5R r - GMm R E P

22 Graph of TE Thus a spacecraft must SLOW DOWN in order to reach a higher orbit! Graph the total energy TE vs. the radius of orbit and include both KE and PE. SOLUTION: total energy of an orbiting satellite + GMm 2R - GMm 2R - GMm R R 2R 3R 4R 5R FYI Kinetic energy (thus v) DECREASES with radius. r E K E E P

23 TE of a satellite is always negative Remember: Friction causes TE to decrease R decreases As R, v As v, KE When would TE be zero? Far from gravitational pull so that PE = 0 and if motionless, KE = 0, hence TE = 0 If KE = PE the object has just enough energy to move to distance where PE = 0, but no KE would be left motionless

24 Conclusion If TE = 0 can just escape from central object If TE > 0 can escape and keep going If TE < 0 cannot escape and is bound to planet. This is called binding energy BE The extra energy needed to free the object is the BE. BE = - TE = - ( - GMm 2R ) = GMm 2R

25 KE = TE = BE Summary PE = 2 KE = 2 TE = 2 BE For a satellite close to a planet R orbit = R planet = R KE = GMm 2R On planets surface BE to break free is BE = GMm 2R total energy of an orbiting satellite

26 Example Calculate the minimum energy required to put a satellite of mass 500kg into an orbit that is at a height equal to the Earth s radius above the surface of the Earth. Solution V E = g 0 R E and V at R = g 0R E 2 V = g 0 R E - g 0R E 2 R = 2 R E R R = g 0 R E 1 R E R V = g 0 R E 1 R E 2 R E = g 0 R E W = Vm = g 0R E m 2 = 10x6.4x106 x = g 0R E 2 = 16000MJ In orbit it will have KE, hence TE = PE + KE = x g 0R 2 E m = x g 0 R E 2 m = x x6.4x106 x 500 = 24000MJ 4R E 4 2R

27 Example KE is POSITIVE and decreasing. GPE is NEGATIVE and increasing (becoming less negative).

28 Example From Kepler s 3 rd law, T 2 = [ 4 2 / (GM) ] r 3. Thus r 3 = [ GM / (4 2 ) ]T 2. That is to say, r 3 T 2.

29 Example From Kepler s 3 rd law T 2 = [ 4 2 / GM ]r 3. Then T = { [ 4 2 /GM ]r 3 } 1/2 T = [ 4 2 / GM ] 1/2 r 3/2 T r 3/2.

30 Example From Kepler s 3 rd law T X 2 = (4 2 / GM)r X3. From Kepler s 3 rd law T Y 2 = (4 2 / GM)r Y3. T X = 8T Y T X 2 = 64T Y2. T X 2 / T Y 2 = (4 2 / GM)r X 3 / [(4 2 / GM)r Y3 ] 64T 2 Y / T 2 Y = r 3 X / r 3 Y 64 = (r X / r Y ) 3 r X / r Y = 64 1/3 = 4

31 Example Since the satellite is in uniform circular motion at a radius r and a speed v, it must be undergoing a centripetal acceleration. Since gravitational field strength g is the acceleration, g = v 2 / r.

32 Example R x F = mg = GMm / x 2 = mv 2 / x. Thus v 2 = GM / x. Finally v = GM / x.

33 continued R x From (a), v 2 = GM / x. But E K = (1/2)mv 2. Thus E K = (1/2)mv 2 = (1/2)m(GM / x) = GMm / (2x). E P = mv and V = GM / x. Then E P = m( GM / x) = GMm / x.

34 continued E = E K + E P E = GMm / (2x) + GMm / x [ from (b)(i) ] E = 1GMm / (2x) + - 2GMm / (2x) E = GMm / (2x). R x

35 continued The satellite will begin to lose some of its mechanical energy in the form of heat. R x

36 continued Refer to E = GMm / (2x) [ from (b)(ii) ]. If E, then x (to make E more negative). If r the atmosphere gets thicker and more resistive. Clearly the orbit will continue to decay (shrink). R x

37 Topic 10: Fields - AHL 10.2 Fields at work Orbital motion, orbital speed and orbital energy E = GMm / (2r) E K = GMm / (2r) E P = GMm / r If r decreases E K gets bigger. If r decreases E P gets more negative (smaller). total energy of an orbiting satellite

38 Example Escape speed is the minimum speed needed to travel from the surface of a planet to infinity. It has the formula v esc 2 = 2GM / R.

39 continued To escape we need v esc 2 = 2GM / R e. The kinetic energy alone must then be E = (1/2)mv esc 2 = (1/2)m(2GM / R e ) = GMm / R e. This is to say, to escape E = 4GMm / (4R e ). Since we only have E = 3GMm / (4R e ) the probe will not make it into deep space.

40 continued Recall that E P = GMm / r. Thus E P = GMm ( 1 / R 1 / R e ).

41 continued The probe is in circular motion so F c = mv 2 / R. But F G = GMm / R 2 = F c. Thus mv 2 / R = GMm / R 2 or mv 2 = GMm / R. Finally E K = (1/2)mv 2 = GMm / (2R).

42 continued The energy given to the probe is stored in potential and kinetic energy. Thus E K + E P = E GMm / (2R) GMm(1/ R 1/ R e ) = 3GMm / (4R e ) 1 / (2R) 1 / R + 1 / R e = 3 / (4R e ) 1 / (4R e ) = 1 / (2R) R = 2R e.

43 Example It is the work done per unit mass by the gravitational field in bringing a small mass from infinity to that point. COMPARE: The work done by the gravitational field in bringing a small mass from infinity to that point is called the gravitational potential energy. The phrase only differs by omission of per unit mass.

44 continued V = GM / r so that V 0 = GM / R 0. But g 0 R 0 = (GM / R 02 )R 0 = GM / R 0 = V 0. Thus V 0 = g 0 R 0.

45 = = R 0. At R 0 = clearly V 0 = From previous problem g 0 = V 0 / R 0 = / = 8.0 m s -2.

46 continued This solution assumes probe is not in orbit but merely reaches altitude (and returns). V g = ( ) 10 7 = E K = E P E K E K0 = E P (1/2)mv 2 = E P v 2 = 2 E P / m v 2 = 2 V g v 2 = 2( ) v = 8000 ms -1.