# Solving Gravitational Force to Exceed Earth Pull on Spaceship

• aal0315
In summary, to find the distance from the Earth at which the pull of the Moon on a spaceship exceeds the pull of the Earth, you need to set the force equations for the Earth and the Moon equal to each other. This can be done by using the average distance between the Earth and the Moon, R, and the distance from the Earth to the spaceship, r. Solving for r using the quadratic equation, the distance is approximately 1.56x10^7 km.
aal0315

## Homework Statement

A spaceship is launched and starts moving directly towards the Moon. At what distance from the Earth will the pull of the Moon, on the spaceship, excced the pull of the Earth? Ignore the effect of the sun in this calculation.

F=G(M1M2/r^2)

## The Attempt at a Solution

i don't know where to begin. does the force of the Earth on the spaceship = the force of the moon on the spaceship and i solve for r?

aal0315 said:

## Homework Statement

A spaceship is launched and starts moving directly towards the Moon. At what distance from the Earth will the pull of the Moon, on the spaceship, excced the pull of the Earth? Ignore the effect of the sun in this calculation.

F=G(M1M2/r^2)

## The Attempt at a Solution

i don't know where to begin. does the force of the Earth on the spaceship = the force of the moon on the spaceship and i solve for r?

You're on the right track. You want to find the distance from the Earth at which the force on the spaceship by the Earth equals the force on the spaceship by the moon. You will need to write the correct force law for each case. That is one force law with Earth and spaeship masses and one with moon and spaceship masses. The you will have to figure out how to represent the distance so that you have only one variable to solve for when you set those two forces equal. I'd let R = the average distance between the Earth and the moon (you can look it up) and r the distance from the Earth to the space ship. Then what is the distance from the spaceship to the moon? Set up your force equations, set them equal, watch stuff cancel out and then solve for r. It will be a quadratic equation so be careful. You may need the quadratic formula. I don't know for sure because I haven't completed the algebra.

You need to break it into two parts.
Force on the rocket from the earth
Force on the rocket from the moon
When these two forces are equal, then the net force acting on the rocket will be zero

To solve this you can write each of the first two parts in terms of the equation you gave, using m1 as the mass of the rocket. For the first equation, m2 would be the mass of the earth, and r the distance from the Earth to the rocket. For the second equation m2 would be the mass of the moon, and r the distance from the rocket to the moon

Set these two equations equal and solve for r to get the point where the spaceship will feel no force. Once beyond this point (closer to the moon), the rocket will feel a stronger pull from the moon.

so i have
G(MshipMearth/r) = G(MshipMmoon/r)
Mearth/r = Mmoon/r (because G and Mship cancel each other out.
but if i try to solve for r, won't they just cancel out as well?

No, because the R is different in each case. The R for the moon case is the distance between the ship and the moon's center of gravity, while the R for the Earth case is the R between the ship and the Earth's center of gravity.

so would it be Mearth / (Rearth + r) = Mmoon / (Rmoon +r)
or Mearth / (distance b/w Earth and moon - r) = Mmoon / (distance b/w Earth and moon - r)?
i know that the distance between the Earth and the moon is 384x10^3 km

i don't know what I am doing at all. any help would be great.

aal0315 said:
so would it be Mearth / (Rearth + r) = Mmoon / (Rmoon +r)
or Mearth / (distance b/w Earth and moon - r) = Mmoon / (distance b/w Earth and moon - r)?
i know that the distance between the Earth and the moon is 384x10^3 km

i don't know what I am doing at all. any help would be great.

Go back to my post. I told you how to write the r in each case. Again, (1) let the erath-spaceship distance be r. (2) let the moon-spaceship distance be (R-r) where capital R is the average earth-moon distance. Plug those into the expression for the force of gravity. That is the only way you will end up with an equation in 1 unknown --the r that you want.

Sounds right?

think I got the calculations wrong. Now I got 3.79x10^5

so the above answers are wrong because i know i never squared the r and R-r
i have tried the calculations over and over and get this huge number that is bigger than the distance of the Earth to the moon. can someone help me?
Me(R^2) - (Me)(2)(R)(r) - (Me)(r^2) = Mm (r^2)
(5.98x10^24)(384x10^6)-2(5.98x10^24)(384x10^6)r-(5.98x10^24)r^2=7.35x10^22 r^2
8.82x10^41 - 4.59x10^33r + 5.98x10r^2 = 7.35x10r^2
4.59x10^33 ± √[[(4.59x10^33)^2 - 4(5.91x10^24)(8.82x10^41)]/2(5.91x10^24)]
i get 4.59x10^33 which is too big. can anyone help please

## What is gravitational force?

Gravitational force is the attractive force between two objects with mass. It is responsible for the movement of planets around the sun and other celestial bodies.

## Why is it important to solve gravitational force on a spaceship?

Knowing how to solve gravitational force on a spaceship is crucial for space exploration and travel. It allows us to understand how objects will behave in space and how to navigate and control spacecrafts.

## How do you calculate gravitational force?

Gravitational force can be calculated using Newton's Law of Universal Gravitation, which states that the force between two objects is directly proportional to the product of their masses and inversely proportional to the square of the distance between them. The formula is F = G * (m1 * m2 / r^2), where F is the force, G is the gravitational constant, m1 and m2 are the masses of the two objects, and r is the distance between them.

## What factors affect gravitational force?

The two main factors that affect gravitational force are the masses of the objects and the distance between them. The larger the masses, the greater the force, and the closer the objects are, the stronger the force will be.

## Can gravitational force be exceeded?

Yes, gravitational force can be exceeded by using a powerful enough force, such as rocket propulsion, to counteract the force of gravity. This allows spacecrafts to leave Earth's gravitational pull and travel through space.

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