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

In summary: This is because M_s is the mass of the ship, and it is the same ship at both sides of the equation. This is true in general. So you are left with:M_e / (R_es)^2 = M_m / (R_ms)^2
  • #1
ML1
6
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.
 
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  • #2
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:
  • #3
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
 
  • #4
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.
 
  • #5


I would like to clarify that the concept of universal gravitation is a fundamental law of physics that explains the attraction between objects with mass. It states that every object in the universe attracts every other object with a force that is directly proportional to the product of their masses and inversely proportional to the square of the distance between them.

In regards to the homework statement, it is important to note that the gravitational pull of an object is not zero, even at the point where two forces are equal. This is because gravity is a continuous force that decreases with distance, but never reaches zero.

To determine the location where the gravitational pull of the Earth and Moon are equal, we can use the equation provided, where F_G represents the force of gravity, G is the universal gravitational constant, m_1 and m_2 are the masses of the Earth and Moon, and r is the distance between them.

We can rearrange the equation to solve for r, which would give us the distance at which the forces are equal. However, this distance would constantly change as the spaceship moves closer to the Moon. Therefore, it would be more accurate to award the certificate at the point where the forces are closest to being equal, rather than at a specific distance.

In conclusion, the certificate should be awarded at the point where the spaceship is closest to the Moon, as this would be the location where the gravitational pull of the Earth and Moon are most balanced.
 

1. What is Universal Gravitation?

Universal gravitation is a fundamental law of physics that describes the force of attraction between two objects with mass. It states that every object in the universe exerts a force on every other object, and the strength of this force depends on the mass and distance between the objects.

2. Who discovered Universal Gravitation?

Isaac Newton is credited with discovering the law of universal gravitation in 1687. He developed the theory based on his observations of the motion of objects and the laws of motion.

3. How does Universal Gravitation affect the motion of planets?

Universal gravitation is responsible for keeping planets in orbit around the sun. The gravitational force between the sun and the planets keeps them in a constant state of motion, balancing the outward force of the planet's velocity and the inward force of the sun's gravity.

4. How is Universal Gravitation related to the force of gravity?

The force of gravity is a specific case of the more general law of universal gravitation. It describes the specific force of attraction between two objects with mass, such as the Earth and a person standing on its surface. The law of universal gravitation encompasses the force of gravity and extends it to all objects in the universe.

5. Can Universal Gravitation be seen in everyday life?

Yes, Universal Gravitation can be seen in everyday life, from objects falling to the ground to the movements of celestial bodies. It is also responsible for keeping our solar system and galaxy in order. Without the force of gravity, the universe would not exist as we know it.

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