Question

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I need help with PART B
It doesn't seem so complicated since it just conservation of momentum and energy but I'm missing the big picture. How should I set up the initial angular moment/energy and final angular momentum/energy?

Answer 1

Even I am trying. Let me go step by step and stop me wherever you have a doubt. We will solve it together

Answer 2

First let me assume r_p -->radius at perihilion
r_a -->radius at aphilion

Answer 3

v_p -->velocity at perihilion
v_a -->velocity at aphilion

Answer 4

Now, We know that angular momentum(L) is conserved

L_p = (r_p)*(m)*(v_p)
L_a = (r_a)*(m)*(v_a)

L_p = L_a

So we get

(r_p)*(m)*(v_p) = (r_a)*(m)*(v_a)
(r_p)(v_p)=(r_a)(v_a)

Answer 5

@Dumboy , understoood until here ??

Answer 6

yeah I got to that far but I'm stuck how I should combine that with energy equation..

Answer 7

Good,

Now we must conserve energy.

m_s → mass of sun
m_sp -->mass of spacecraft


Total energy at perihelion
= Kinetic energy(K.E) of spacecraft + gravitational energy between spacecraft and sun
= ½ * m* (v_p)^2 + G*(m_s) *(m_sp) / (r_p)^2


Total energy at aphilion
= Kinetic energy(K.E) of spacecraft + gravitational energy between spacecraft and sun
= ½ * m* (v_a)^2 + G*(m_s) *(m_sp) / (r_a)^2

Answer 8

m = m_sp

Answer 9

@Dumboy , any problems/doubt until here ??

Answer 10

are m_sp going to cancel out in the end? cause it's value is not given

Answer 11

yes :) You will see now

Answer 12

since total energy is conserved, we get
½ * m* (v_p)^2 + G*(m_s) *(m_sp) / (r_p)^2
= ½ * m* (v_a)^2 + G*(m_s) *(m_sp) / (r_a)^2

½ * m* ( (v_p)^2 – (v_a)^2)
= G*(m_s) *(m_sp) ( 1 / (r_a)^2 - 1/(r_p)^2 )

Since m = m_sp.

½ * ( (v_p)^2 – (v_a)^2) = G*(m_s) * ( 1 / (r_a)^2 - 1/(r_p)^2 )

Answer 13

@Dumboy , any problems/doubt until here ??

Answer 14

nope

Answer 15

Now
use this relation we derived above from conservation of angular momentum
(r_p)(v_p)=(r_a)(v_a)

Therefore,
(r_p)=(r_a)(v_a) / (v_p)

Substitute this r_p in the equation by conservation of energy relation and tell me what you you get??

Answer 16

shouldn't we substitute one of the v's so we can get rid of one v?

Answer 17

Yes. Exactly that is what I was thinking. That's why I asked you to think from here

Answer 18

Good point :)

Answer 19

Yes you continue from here.. and tell me what you get?

Answer 20

I am facing a problem. I am getting square root of negative number. Same problem or you are getting the right answer??

Answer 21

I am getting v_p = sqrt(- 2*G*m_s)/r_p

Answer 22

??

Answer 23

I got to \[\sqrt{(2Gm_sr_p^2-2Gm_sr_a^2)/(r_p^2(r_a^2-r_p^2))}\]
I'm sure how I could simplify this further..

Answer 24

not sure*

Answer 25

Take 2G_s common outstide. and cancel numerator and denominator but with a minus sign. You will get same answer as I got :(

Answer 26

@mos1635 , any help bro..

Answer 27

I meant you can provide any help :P

Answer 28

what do you mean by cancel numerator and denominator but with a minus sign?

Answer 29

\[\sqrt{2*G*m_S (r_p^2 -r_a^2)/ r_p^2(r_a^2 -r_p^2)}\]
\[\sqrt{- 2*G*m_S/r_p^2 }\]

Answer 30

oh I see..
can you say G is -G since gravity acts downward and cancel out the negatives?

Answer 31

no way. There is some conceptual mistake
@mos1635 , @apoorvk @Vincent-Lyon.Fr

Answer 32

@experimentX

Answer 33

gravitonial potensial energy does not need r^2 but r

Answer 34

Hi! The method described by shivam_bhalla is correct, but there is a mistake in the expression of the gravitational potential energy.

GPE is not:
GPE = + G*(m_s) *(m_sp) / (r_p)^2
but is (with minus sign):
GPE = - G*(m_s) *(m_sp) / (r_p) <------- not squared

Answer 35

hawk eye @mos1635, and @Vincent-Lyon.Fr

Answer 36

@Dumboy , corect the mistake and check you are getting the right answer. I am also checking :)

Answer 37

I am getting it :)
@Dumboy , hope you are getting the final answer too.
@mos1635 and @Vincent-Lyon.Fr --> thanks for pointing out my blunder :P

Answer 38

you did all the work.nice job!!

Answer 39

I checked and it leads to the answers given in the problem.

Answer 40

Thanks :D. @Dumboy , please give the medal to @Vincent-Lyon.Fr . I cannot give more than 1 medal.
@Dumboy , if you are having any problem, just tag me here. :)

Answer 41

yeah I got the answer too
thank you for the help everyone :)

Answer 42

Welcome :)

Answer 43

Welcome!