A Rocket's Distance from the center of the moon

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SUMMARY

The discussion centers on calculating the maximum distance a projectile can reach from the center of the Moon after being launched with an initial speed of 500 m/s. The relevant equations include energy conservation principles, specifically the kinetic energy equation (K = 0.5mv²) and gravitational potential energy (U = -GMm/r). The correct maximum distance, as confirmed by the answer key, is 1.84E6 m, achieved by applying the equation r = (-2*M_e*G*R_e)/(v²*R_e - 2*G*M_e). Participants emphasized the importance of correctly identifying the signs of potential energy in the calculations.

PREREQUISITES
  • Understanding of gravitational potential energy and kinetic energy concepts.
  • Familiarity with the gravitational constant (G = 6.7E-11 m³/kg·s²).
  • Knowledge of energy conservation principles in physics.
  • Ability to manipulate algebraic equations for solving physics problems.
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  • Learn about gravitational potential energy and its role in orbital mechanics.
  • Explore advanced topics in classical mechanics, focusing on energy conservation.
  • Practice solving similar problems involving energy conservation in gravitational fields.
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This discussion is beneficial for physics students, educators, and anyone interested in understanding projectile motion and gravitational interactions, particularly in the context of celestial bodies like the Moon.

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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...
 
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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.
 
I didn't get the right answer...
 
Show your calculations.
 
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
 
So is there something I am missing I thought a simple energy conservation would cover it
 
Use the equation :

K+U = K_o+U_o

where K could be 1/2mv^2
and U = -GMm/r
 
tnutty said:
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
 
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
Could I get a short explanation on that? I don't follow your variables
 
  • #11
gc33550 said:
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
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
tnutty said:
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 don't know if anyone is going to do well on it, its bad if I am aiming for a 50% on it...
 
  • #14
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
I also have my final today. Your's not with jones is it?
 
  • #16
Jones? I go to Purdue
 
  • #17
By the way I got it thanks guys... Those negative signs will get you every time haha
 
  • #18
NO problem
 

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