# Net gravitational force on an object

• Jaccobtw
In summary, the conversation discusses finding the net x and y components of the gravitational force between two objects with given masses and distances, and how to express these forces using direction vectors and unit vectors. The resultant force can be calculated by summing the individual forces using their respective vectors.
Jaccobtw
Homework Statement
A test object of mass m is placed at the origin of a two dimensional coordinate system. An object 1, of the same mass, is at (d, 0), and an object 2, of mass 2m, is at (-d, l). What is the magnitude of the vector sum of the gravitational forces exerted on the test object by the other two objects?
Relevant Equations
F(gravity) = G * (M * m)/ (r^2)
G = 6.6738 * 10^-11 N * m^2

I suppose we can just find the net x components and y components and then go from there.
Σ Fx = F(mass 1) - Fx(mass 2)
G* (m^2)./d^2) - something

I'm not sure how to express the component forces of the 2nd mass

Object 1 is at (d, 0). The distance from object 0 to object 1 is r=d. The direction vector from object 1 to object 0 is <-d, 0> the unit vector in that direction is <-1, 0>. The magnitude of the gravitational force object 1 exert on object 0 is $\frac{Gm^2}{d^2}$ and the vector force is $\frac{Gm^2}{d^2}\left<-1, 0\right>$. Since you are asking about object 2 I presume you got that.

Object 2 is at (-d, l). The distance from object 0 to object 2 is $r=\sqrt{d^2+ l^2}$. The direction vector from object 2 to object 0 is <d, -l>. The magnitude of the gravitational force object 1 exerts on object 0 is $\frac{2Gm^2}{d^2+ l^2}$ and the vector force is $\frac{2Gm^2}{d^2+ l^2}\left<d, -l\right>$.

The resultant is the sum of those, $\left<\frac{-Gm^2}{d^2}, 0\right>+ \left<\frac{2dGm^2}{d^2+ l^2}, \frac{-2lGm^2}{d^2+ l^2}\right>$.

Jaccobtw
HallsofIvy said:
The resultant is the sum of those, $\left<\frac{-Gm^2}{d^2}, 0\right>+ \left<\frac{2dGm^2}{d^2+ l^2}, \frac{-2lGm^2}{d^2+ l^2}\right>$.
Of course you meant to write
"The resultant is the sum of those, $\left<\frac{-Gm^2}{d^2}, 0\right>+ \left<\frac{2dGm^2}{(d^2+ l^2)^{3/2}}, \frac{-2lGm^2}{(d^2+ l^2)^{3/2}}\right>$."

## 1. What is net gravitational force on an object?

Net gravitational force on an object is the combined force of all gravitational forces acting on the object. It is the sum of all the attractive forces between the object and other objects in its surroundings.

## 2. How is net gravitational force calculated?

Net gravitational force is calculated by multiplying the mass of the object by the acceleration due to gravity (9.8 m/s²) and the direction of the force, which is always towards the center of the Earth. This can be represented by the formula F = mg, where F is the net gravitational force, m is the mass of the object, and g is the acceleration due to gravity.

## 3. How does distance affect net gravitational force?

The farther an object is from another object, the weaker the gravitational force between them. This is because the force of gravity decreases with distance according to the inverse square law. This means that the force decreases as the square of the distance between the objects increases.

## 4. What factors can affect the net gravitational force on an object?

The net gravitational force on an object can be affected by the masses of the objects involved, their distance apart, and the presence of other objects that may exert gravitational force on the object in question. The shape and composition of the objects can also have an impact on the net gravitational force.

## 5. How does net gravitational force differ from weight?

Net gravitational force is the total force of all gravitational forces acting on an object, while weight is the measure of the force of gravity on an object's mass. Weight can vary depending on the strength of the gravitational field it is in, while net gravitational force remains constant regardless of the location of the object.

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