Where Should the Certificate Be Awarded on a Space Tour from Earth to Moon?

Click For Summary

Homework Help Overview

The discussion revolves around determining the appropriate point in a space tour from Earth to the Moon where a certificate should be awarded to participants. The context involves gravitational forces acting on a spaceship as it travels between the two celestial bodies.

Discussion Character

  • Conceptual clarification, Mathematical reasoning, Assumption checking

Approaches and Questions Raised

  • Participants explore the relationship between gravitational forces from Earth and the Moon, questioning how to set up the equations correctly. There is confusion regarding the conditions under which the forces balance and how to express the distances involved.

Discussion Status

Some participants have offered guidance on setting up the gravitational force equations and have clarified the relationship between the distances from the Earth and Moon to the spaceship. There is ongoing exploration of the mathematical relationships and assumptions involved.

Contextual Notes

Participants are navigating the complexities of gravitational forces and their magnitudes, with specific attention to the scenario where the spaceship is aligned between the Earth and the Moon. There is a noted confusion regarding the implications of setting forces equal to zero and the relevance of the distances involved.

ML1
Messages
6
Reaction score
0

Homework Statement



Transplanertay tours promises tour participants a certificate to commemorate their passage from the stronger influence of Earth's graviational pull to the stronger pull of the Moon at the point where the two forces on your spaceship are equal. Where on the trip should you award the certificate?

Homework Equations



F_G = G (m_1)(m_2)/ (r^2)

The Attempt at a Solution



We know that F will equal zero right...? I'm confused how to go about it. I know I'm solving for r but if F= 0 i can't solve for anything.
 
Physics news on Phys.org
ML1 said:
We know that F will equal zero right...?
Yes. :approve: [Edit: Well, at least the component of the resulting force that is parallel to the line intersecting the Earth and the Moon will be zero. If the ship is not on that line, there will be non-zero component of the net force pushing it toward that line.]
I'm confused how to go about it. I know I'm solving for r but if F= 0 i can't solve for anything.
There are two rs involved. There is the distance from the center of the Earth to the ship rearth_ship, and the distance from the ship to the center of the Moon rship_moon.

If the ship happens to be on a line intersecting the Earth and Moon, These two distances are related by the distance from the center of the Earth to the center of the Moon dearth_moon.

dearth_moon = rearth_ship + rship_moon *​

*(Again though, the above equation is a special case where the ship is on a straight line between the Earth and the Moon. If you keep things in terms of both rearth_ship and rship_moon, you don't need to use the above equation, and your answer will apply more generally.)

In general, there are two forces on the ship we are concerned with. There is the force of gravity from the Earth and there is the force of gravity from the Moon. It is the sum of these equal and opposite force magnitudes that equals zero -- not just anyone particular magnitude.

[Edit: in other words, set the magnitude of the Earth's gravitational force on the ship equal to the Moon's gravitational force magnitude on the ship, and simplify.]
 
Last edited:
collinsmark said:
[Edit: in other words, set the magnitude of the Earth's gravitational force on the ship equal to the Moon's gravitational force magnitude on the ship, and simplify.]

Like this...?

F (M_e)(M_s) / (R_es)^2 = F (M_s)(M_m) / (R_ms)^2 ?

:confused:

es - Earth/ship

ms - moon shop
 
ML1 said:
Like this...?

F (M_e)(M_s) / (R_es)^2 = F (M_s)(M_m) / (R_ms)^2 ?

:confused:

es - Earth/ship

ms - moon shop
Yes, except remove the "F" from both sides of the equation.

You can also simply further too. Notice that M_s cancels out.
 

Similar threads

Replies
21
Views
3K
  • · Replies 1 ·
Replies
1
Views
2K
Replies
2
Views
2K
Replies
2
Views
3K
  • · Replies 73 ·
3
Replies
73
Views
6K
  • · Replies 1 ·
Replies
1
Views
1K
  • · Replies 5 ·
Replies
5
Views
3K
Replies
5
Views
3K
Replies
1
Views
2K
  • · Replies 2 ·
Replies
2
Views
3K