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

AI Thread Summary
The discussion centers on determining the optimal point to award a certificate during a space tour from Earth to the Moon, specifically at the location where gravitational forces from both celestial bodies are equal. Participants express confusion over the calculations needed to find this point, particularly regarding the gravitational force equations. It is clarified that the forces acting on the spaceship from Earth and the Moon can be equated to find the balance point, which involves understanding the distances from each body to the ship. The conversation emphasizes that the gravitational force magnitudes must be equal for the net force to be zero, leading to the conclusion that the certificate should be awarded at this equilibrium point. Overall, the focus is on the physics of gravitational forces in space travel.
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.
 
Kindly see the attached pdf. My attempt to solve it, is in it. I'm wondering if my solution is right. My idea is this: At any point of time, the ball may be assumed to be at an incline which is at an angle of θ(kindly see both the pics in the pdf file). The value of θ will continuously change and so will the value of friction. I'm not able to figure out, why my solution is wrong, if it is wrong .
Thread 'Voltmeter readings for this circuit with switches'
TL;DR Summary: I would like to know the voltmeter readings on the two resistors separately in the picture in the following cases , When one of the keys is closed When both of them are opened (Knowing that the battery has negligible internal resistance) My thoughts for the first case , one of them must be 12 volt while the other is 0 The second case we'll I think both voltmeter readings should be 12 volt since they are both parallel to the battery and they involve the key within what the...
Back
Top