# What is the Distance from Earth to the Moon where Gravitational Pull is Equal?

• UrbanXrisis
In summary, the distance from Earth to the moon where the gravitational pull of the moon is equal to the gravitational pull of the earth can be calculated by using the equation F_e = F_m, where F is the force of gravity, G is the gravitational constant, M is the mass, and r is the distance between the two objects. After solving the equation and graphing the function, the distance was found to be 3.84 x 10^8 meters. Another method is to plot the acceleration due to gravity for both Earth and the moon and find the intersect point, which would also give the same result.
UrbanXrisis
anyone know the distance from Earth (to the moon) where the gravitational pull of the moon is equal to the gravitational pull of the earth? (or any way I could calculate it?)

Last edited:
A point between the two? Just use the same method you used in the other post. It will work for this example.

I tried, I got a nonreal number

Last edited by a moderator:
384,400,000meters between Earth and moon

earth mass is 5.98 x 10^24
moon mass is 7.35 x 10^22

$$F_e = F_m$$

$$F = G\frac{Mm}{r^2}$$

$$\frac{5.98x10^{24}}{x^2} = \frac{7.35x10^{22}}{y^2}$$

$$x + y = 3.84 x 10^8$$

You don't need to add the radius of the Earth and moon on the fractions on the topright. You can regard them as point masses. Solving the system above should get you the correct answer.

Does this distance takes account of the radius of both?

UrbanXrisis,

First of all, NEVER, EVER, EVER start plugging in numbers until you have solved the equation for the variable you need. It wastes time, and it creates countless chances for errors.

Second, it looks to me as though you're using the radius of the Earth and moon in the denominators of your law of gravity equations. Why? How is the distance r in the denominator defined?

I got x to equal something between 3 and 4 when I graphed the function...

Last edited by a moderator:
$$\frac{5.98x10^{24}}{x^2} = \frac{7.35x10^{22}}{y^2}$$

$$M_ey^2 = M_mx^2$$

$$x+y = r$$

$$M_e(r-x)^2 = M_mx^2$$

$$\frac{M_m}{M_e} = \left(\frac{r-x}{x}\right)^2$$

$$\sqrt{\frac{M_m}{M_e}}x = r-x$$

$$\sqrt{\frac{M_m}{M_e}}x+x = r$$

$$(\sqrt{\frac{M_m}{M_e}}+1)x = r$$

$$\frac{r}{\left(\sqrt{\frac{M_m}{M_e}}+1\right)} = x$$

$$r = 3.84 x 10^8m$$

A much easier way would be to realize that:

$$a = \frac{GM}{r^2}$$

and plot a for Earth and a for the moon, and find the intersect point.

## What is equal gravitational pull?

Equal gravitational pull is a physical phenomenon in which two or more objects exert the same amount of gravitational force on each other. This means that the force of gravity between the objects is balanced and there is no net acceleration.

## How is equal gravitational pull measured?

Equal gravitational pull can be measured using the universal law of gravitation, which states that the force of gravity between two objects is directly proportional to the product of their masses and inversely proportional to the square of the distance between them.

## What are some examples of equal gravitational pull?

Some examples of equal gravitational pull include the Earth and the Moon, the Sun and the Earth, and two objects of equal mass that are positioned at the same distance from each other.

## What factors affect equal gravitational pull?

The factors that affect equal gravitational pull include the masses of the objects and the distance between them. The greater the masses of the objects, the stronger the gravitational pull, while the greater the distance between them, the weaker the gravitational pull.

## Why is equal gravitational pull important?

Equal gravitational pull is important because it helps to keep objects in orbit and maintain the balance of the universe. It also allows for the formation and stability of celestial bodies, such as planets and stars.

• Introductory Physics Homework Help
Replies
12
Views
1K
• Introductory Physics Homework Help
Replies
40
Views
1K
• Introductory Physics Homework Help
Replies
21
Views
2K
• Astronomy and Astrophysics
Replies
14
Views
821
• Introductory Physics Homework Help
Replies
28
Views
1K
• Introductory Physics Homework Help
Replies
2
Views
1K
• Introductory Physics Homework Help
Replies
5
Views
193
• Introductory Physics Homework Help
Replies
4
Views
436
• Introductory Physics Homework Help
Replies
5
Views
1K
• Introductory Physics Homework Help
Replies
3
Views
1K