HW6 Physics 311 Mechanics

Transcription

1 HW6 Physics 311 Mechanics Fall 015 Physics depatment Univesity of Wisconsin, Madison Instucto: Pofesso Stefan Westehoff By Nasse M. Abbasi June 1, 016

2 Contents 0.1 Poblem Poblem Poblem Pat (1) Pat () Poblem Poblem Pat (1) Pat ()

3 Fall 015 Homewok 6 (10/3/15, due 10/30/15) 0.1 Poblem 1 1. (5 points) An Eath satellite has a speed of 8,070km/h when it is at its peigee of 0km above Eath s suface. Find the apogee distance, its speed at apogee, and its peiod of evolution. SOLUTION: Fom. the (5 points) vis-viva elation A spacecaft is in cicula obit 00 km above ( Eath s suface. What minimum velocity α kick must be applied to let vthe peigee spacecaft = escape 1 ) fom Eath s influence? What is the (1) m spacecaft s escape tajectoy with espect to Eath? p a Whee m is the educed mass and α = GM eath m satt, which educes to GM eath and known constant called the Standad gavitational paamete which fo eath is (Fom table) 3. (15 points) α A comet is obseved to have a speed m = v when it km3 /s is at a distance fom the Sun. Its diection of motion makes an angle φ with the adius vecto fom the Sun. And (1) Find the eccenticity of the comet s obit. p = () If the velocity of the comet is expessed as q times the Eath s velocity and its distance tothesunasdastonomicalunits,showthattheobitofthecometishypebolic,paabolic, = 6598 km Whee o 6378 elliptic, is the depending equatoial on adius whethe of eath. the quantity And v q d is geate than, equal to, o less than, peigee = 8070 km/h. Theefoe, we use (1) to solve fo espectively. a, the length of the semimajo axes of the elliptical obit of the satellite aound the eath. Fom (1), by squaing both sides vp = α ( 1 ) m p a ( ) ( 8070 = ) a Solving fo a gives Hence the apogee distance is We can also find a = 6640 km a = 1380 km a = a p = = 668 km...continued on next page... When the satellite is at the apogee, it will be above the eath at height of h a = a eath = = 304 km 3

4 The peiod T is given by T = π a 3 α m = π = 5385 sec = = h 4

5 1. (5 points) An Eath satellite has a speed of 8,070km/h when it is at its peigee of 0km above Eath s suface. Find the apogee distance, its speed at apogee, and its peiod of evolution. 0. Poblem. (5 points) A spacecaft is in cicula obit 00 km above Eath s suface. What minimum velocity kick must be applied to let the spacecaft escape fom Eath s influence? What is the spacecaft s escape tajectoy with espect to Eath? SOLUTION: 3. (15 points) The total enegy is A comet is obseved to have a speed v when it is at a distance fom the Sun. Its diection of motion makes an angle φ with E = the 1 mṙ adius + Uvecto effective fom the Sun. The escape (1) Find velocity the eccenticity is when U effective of the= comet s 0, theefoe obit. () If the velocity of the comet is expessed as q times the Eath s velocity and its distance tothesunasdastonomicalunits,showthattheobitofthecometishypebolic,paabolic, 0 = U + l m o elliptic, depending on whethe the quantity q d is geate than, equal to, o less than, But angula espectively. momentum l = mv and U = GMem, hence the above becomes 0 = GM em = GM em = GM e + m v m + mv Now we ae given that the satellite was at = = 6578 km (this is p fo the new obit as well). Using GM e = km 3 /s fom tables then we solve now fo v in (1), which will be the new velocity. Hence + v 0 = v v = km/sec Befoe this, the spacecaft was in cicula obit. So its speed was (1)...continued on next page... α 1 v c = m = 6578 = km/sec The diffeence is the minimum speed kick needed, which is = 3.5 km/sec This obit is paabolic since U effective = 0 as seen on the U effective vs. gaph. paabolic is the fist obit beyond elliptic that do not contain tun points. The next obit is hypebolic. 5

6 Eath s suface. Find the apogee distance, its speed at apogee, and its peiod of evolution. 0.3 Poblem 3. (5 points) A spacecaft is in cicula obit 00 km above Eath s suface. What minimum velocity kick must be applied to let the spacecaft escape fom Eath s influence? What is the spacecaft s escape tajectoy with espect to Eath? 3. (15 points) A comet is obseved to have a speed v when it is at a distance fom the Sun. Its diection of motion makes an angle φ with the adius vecto fom the Sun. (1) Find the eccenticity of the comet s obit. () If the velocity of the comet is expessed as q times the Eath s velocity and its distance tothesunasdastonomicalunits,showthattheobitofthecometishypebolic,paabolic, o elliptic, depending on whethe the quantity q d is geate than, equal to, o less than, espectively. SOLUTION:...continued on next page Pat (1) Eccenticity is defined as (fo all conic sections) e = Whee α = GM sun m and l is the angula momentum Theefoe (1) becomes e = 1 + El mα (1) l = m v = mv sin φ 1 + E (v sin φ) m (GM sun ) The enegy of the comet is given by E = 1 mv GMsunm, then the above becomes e = 1 + ( ) 1 mv GMsunm (v sin φ) m (GM sun ) ( ( = 1 )) (v 1 + mv GMsunm ) sin φ m = 1 + ( v GM ) ( sun v sin φ 6 GM sun GM sun )

7 0.3. Pat () Let v = qv e whee v e is eath velocity aound the sun and let = d e whee e is the astonomical unit (the distance between the eath and sun) then esult of pat (1) becomes ( e = 1 + (qv e ) GM ) ( ) sun de qv e sin φ () d e GM sun Looking at the eath/sun system, we know that GM sun m eath e GM sun e = m eathv e e = v e GM sun = e v e Replacing GM sun in () by the above esult gives ( e = 1 + (qv e ) ) ( eve de qv e sin φ d e e ve ( ) ( ) = 1 + (qv e ) v e dq sin φ d v e ( = 1 + q ) (dq sin φ) d ( ) q d = 1 + (dq sin φ) d We ae now eady ( to ) answe the final pat. If q d = then e = 1 which means it is paabolic. If q q d > then d is positive and the expession inside. is lage than one, and hence e > 1, d ( ) which means the obit is hypebolic. Finally, if q q d < then d is negative, and the expession d inside. is less than one, which means e < 1 and hence the obit is elliptic. ) 7

8 0.4 Poblem 4 4. (10 points) If the minimum and maximum velocities of a moon otating aound a planet ae v min = v v 0 and v max = v +v 0, show that the eccenticity is given by e = v 0 v. SOLUTION: The angula momentum l is constant. At peigee, whee the speed is maximum, we have 5. (15 points) When a spacecaft is placed into geosynchonous l p = mv max p obit, it is fist launched, along with a populsion stage, into a nea cicula low Eath obit (LEO) using a booste ocket. Then And atthe apogee, populsion whee the stage speed is fied is minimum, and the spacecaft we have is tansfeed to an elliptical tansfe obit designed to take it to geosynchonous altitude at obital apogee. At apogee, the l populsion stage is fied again to take a = mv it out min of the a elliptical obit back into a cicula (now Since l geosynchonous) is constant, then obit. (1) Calculate the equied velocity boost v 1 to move the satellite fom its cicula low Eath obit into the elliptical tansfe mv max obit. p = mv min a () Calculate the equied velocity v max boost p = v v min a to move the satellite fom the elliptical (1) tansfe obit into the geosynchonous cicula obit. But Hence (1) becomes a = a (1 + e) p = a (1 e) v max a (1 e) = v min a (1 + e) v max (1 e) = v min (1 + e) Replacing v max = v + v 0 and v min = v v 0 gives v max ev max = v min + ev min v max v min = e (v min + v max ) e = v max v min v min + v max e = (v + v 0) (v v 0 ) (v + v 0 ) + (v v 0 ) = v 0 v = v 0 v 8

9 If the minimum and maximum velocities of a moon otating aound a planet ae v min = v v 0 and v max = v +v 0, show that the eccenticity is given by 0.5 Poblem 5 e = v 0 v. 5. (15 points) When a spacecaft is placed into geosynchonous obit, it is fist launched, along with a populsion stage, into a nea cicula low Eath obit (LEO) using a booste ocket. Then the populsion stage is fied and the spacecaft is tansfeed to an elliptical tansfe obit designed to take it to geosynchonous altitude at obital apogee. At apogee, the populsion stage is fied again to take it out of the elliptical obit back into a cicula (now geosynchonous) obit. (1) Calculate the equied velocity boost v 1 to move the satellite fom its cicula low Eath obit into the elliptical tansfe obit. () Calculate the equied velocity boost v to move the satellite fom the elliptical tansfe obit into the geosynchonous cicula obit. SOLUTION: 9

10 Velocity at ciula GEO obit Velocity at apogee of ellipse V 3,4 V 4, V 3 R GEO Hohmann tansfe ellipse M E R LEO V 1, V V 1 Velocity in LEO cicula obit Velocity at peigee of ellipse Pat (1) In this calculation, the standad symbol µ is used fo GM eath which is the Standad gavitational paamete (in class, we used α fo this same paamete). Fo eath m µ = km 3 /s The fist step is to find a fo the tansfe ellipse. This is given by a = R LEO + R GEO Next, we fist find V 1, which is velocity in the LEO cicula obit just befoe initial kick to V. Since this is cicula, the speed is given by µ V 1 = R LEO Next step is to find V, which is the speed at the peigee of the ellipse (the tansfe obit). This is given by the standad vis-viva elation ( V = µ 1 ) (1) R LEO a 10

11 Whee R LEO = peigee fo the ellipse. Now that we found V and V 1, then 0.5. Pat () V 1 = V V 1 ( = µ 1 ) µ R LEO a When at the apogee of the tansfe ellipse, the speed is given by ( V 3 = µ 1 ) R GEO a We now want to be of GEO cicula obit, hence µ V 4 = And theefoe, the speed boost is R GEO R LEO V 34 = V 4 V 3 ( µ = µ 1 ) R GEO R GEO a 11