AMM PBL Members: Chin Guan Wei p Huang Pengfei p Lim Wilson p Yap Jun Da p Class: ME/MS803M/AM05

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Cornelius Basil Carroll
5 years ago
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1 AMM PBL Membes: Chin Guan Wei p3674 Huang Pengfei p36783 Lim Wilson p36808 Yap Jun Da p36697 Class: MEMS803MAM05

2 The common values that we use ae: G=6.674 x 0 - m 3 kg - s - Radius of Eath ()= 637km [Fom Univese Today] Mass of Eath (M) = x 0 4 kg [Fom Univese Today] Execise Conside a bullet ocket fied upwads with an initial speed s 0. Is it possible fo the ocket to escape fom the Eath if s 0 is lage enough, o will it always fall back to Eath? Find the escape speed s 0 if possible. It is possible fo the ocket to escape fom the Eath if s 0 is lage enough. The ocket will neve fall back once it escapes fom the Eath as the tavelling distance is infinity because the velocity neve eaches to zeo. Let s be the initial velocity of the bullet c 0 ds d d d s. 5 c 0.5s When, s 0 0.5s s 0 s s ( x 0-0 )() ( x 0 4 ) 637 s.87km s.87km s(3dp)

3 Execise Conside a satellite launched fom a ocket in a diection tangent to the Eath s suface at a height of 00km. i) What should the initial speed of the satellite be fo it to go into a cicula obit aound the Eath? Let be the distance between the satellite and the cente of Eath x cos, y sin t dx dy v, ( sin( t), cos( t)) dt dt d x d y a, ( cos( t), sin( t)) dt dt Since, the satellite is in a cicula obit, the satellite is always a fixed distance fom the cente of eath. Theefoe, the acceleation is constant at all points. The adius() of the cicle is 647km. When t=0, a=(,0) v v a 3 ( x km s km s(3dp) 3 3 )( x 0 4 ) 647

4 Checking using numeical estimation

5 Diagam of the Eath and the path of satellite

6 Execise ii) What is the least initial speed it can have to emain in an obit aound the Eath, i.e. not fall back to Eath? The acceleation of the satellite is not constant hence a elliptic obit will be fomed Let be the distance between the satellite and the cente of Eath Let one of the foci of the ellipse be the cente of the eath Let point be the stating position of the satellite, 00km above Eath Let point be opposite of point, just above the suface of the Eath + = E T ) angula momentum ( consevation of By constant, a mass is Since m m L L m m m m m m m m

7 (3d.p.) 7.89kms 7.885kms Checking with numeical estimation:

8 Execise iii.) What is the geatest initial speed it can have to emain in an obit aound the Eath, i.e. not fly off into the sola system? Fom the law of consevation of enegy. Enegy can neithe be ceated no destoyed, howeve it can convet fom one fom to anothe fom without any enegy loss. By default, the satellite only possesses kinetic and gavitational potential enegy, neglecting othe foms of enegy. Let be the gavitational potential enegy of the satellite. Let be the kinetic enegy of the satellite. Let E T be the total enegy of the satellite. + = E T When the satellite at infinite distance away fom the Eath, the kinetic enegy will become minimum which is closing to zeo. It is due to the speed of the satellite deceasing to the minimum as its escapes infinitely away fom the Eath. To pove that: When 0 E T mv0 Whee Since ( x 0.004km s -0.00km s(3dp) 0 )() ( x 0 Satellite must neve each.kms tostay in obit 4 ) mv0 0 Whee Fom RHS: 0 (change in gavitational potential enegy) = 0 (Fom: Splung.com physics) m( ) Since the only apply when thee is moe than one object. m( ) Fom LHS: mv0 m m m( ) RHS

9 Appoximating with numeical estimation (Plotting 50,000,000 points at sec inteval of t)

10 Bibliogaphy Cain,F.,009.Cicumfeence of the Eath[online].Available fom: [Accessed 0 June 03] Cain,F.,009.Eath s Mass[online]. Available fom: [Accessed 0 June 03] Mihos, C., Obital Enegy[online]. Available fom: [Accessed 7 June 03] Splung.com Physics. Gavitation [online]. Available fom: [Accessed 5 June 03] Wolfam MathWold.Ellipse[online].Available fom: [Accessed 0 June 03] Wolfam MathWold.Paabola[online]. Available fom: [Accessed 0 June 03]

Escape Velocity. GMm ] B

Escape Velocity. GMm ] B 1 PHY2048 Mach 31, 2006 Escape Velocity Newton s law of gavity: F G = Gm 1m 2 2, whee G = 667 10 11 N m 2 /kg 2 2 3 10 10 N m 2 /kg 2 is Newton s Gavitational Constant Useful facts: R E = 6 10 6 m M E