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A Rocket's Distance from the center of the moon

  • Thread starter gc33550
  • Start date
  • #1
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Homework Statement


The Moon has a mass of M= 7.0E22 kg and a radius of R=1.75E6 m. A projectile with
mass m=10 kg is shot straight up from the Moon’s surface with an initial speed of 500 m/s. What is the maximum distance from the center of the Moon that this projectile can reach?

Homework Equations


E_initial=E_final
Kinetic_initial=Grav_potential
.5mv^2=GMm/r


The Attempt at a Solution


Thinking through this I believe this is wrong, because essentially that is the equation for escape velocity. We don't know for sure that this rocket will escape... I am lost...
 

Answers and Replies

  • #2
rl.bhat
Homework Helper
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Thinking through this I believe this is wrong, because essentially that is the equation for escape velocity. We don't know for sure that this rocket will escape... I am lost...

It is true for any velocity. Calculate r.
 
  • #3
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I didn't get the right answer...
 
  • #4
rl.bhat
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Show your calculations.
 
  • #5
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r=(2GM)/v^2 is what i believe I worked it out to so

r=(2*6.7E-11*7E20)/500^2

r=3.752E5

According to the answer key it is supposed to be 1.84E6
 
  • #6
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So is there something I am missing I thought a simple energy conservation would cover it
 
  • #7
327
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Use the equation :

K+U = K_o+U_o

where K could be 1/2mv^2
and U = -GMm/r
 
  • #8
22
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Use the equation :

K+U = K_o+U_o

where K could be 1/2mv^2
and U = -GMm/r
I am assuming that there is no kinetic in my final state because the rocket doesn't have enough velocity to escape according to sqrt(2GM/r) so then:

(GMm/r_final)=.5mv^2+GMm/r_initial

r=GMm/(.5mv^2+(GMm/r_initial))

r= 6.7E-11*7E22*10/(.5*10*500^2+((6.7E-11*7E22*10)/1.75E6))

r=1.672e6
I still don't get the correct answer
 
  • #9
327
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here :

-GMm/ r =1/2m*v^2 - GMm/R_e

r = (1/R_e - v^2 / (2*G*M_e ) ^-1

r = 1.84E^6
 
  • #10
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Could I get a short explanation on that? I don't follow your variables
 
  • #11
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Could I get a short explanation on that? I don't follow your variables
r=(-2*M_e*G*R_e)/(v^2*R_e-2*G*M_e)

r=(-2*7e22*6.7e-11*1.75e6)?(500^2*1.75e6-2*6.7e-11*7e22)

r=1.84e6

Good luck with your test today, I've been up all night studying for it, hope this helps!
 
  • #12
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Maybe my brain is physics fried haha But which Variables need to be negative and why? Is it both potentials? I thought the would have positive potential energy or is it the fact that it is in the negative direction? Oh boy I am going to fail haha
 
  • #13
15
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Use the equation :

K+U = K_o+U_o

where K could be 1/2mv^2
and U = -GMm/r
Use exactly what tnutty has written here for positive and negative values.

Dont worry about the final, I dont know if anyone is going to do well on it, its bad if Im aiming for a 50% on it.....
 
  • #14
LowlyPion
Homework Helper
3,090
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But which Variables need to be negative and why? Is it both potentials? I thought the would have positive potential energy or is it the fact that it is in the negative direction?
The gravitational potential expression is

PE = -GM/r

So for your problem they would both carry a negative sign.
 
  • #15
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I also have my final today. Your's not with jones is it?
 
  • #16
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Jones? I go to Purdue
 
  • #17
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By the way I got it thanks guys... Those negative signs will get you every time haha
 
  • #18
327
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NO problem
 

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