How Does Distance Affect Gravitational Force Between Objects?

[physics1234]

Objects with masses of 285 kg and a 585 kg are separated by 0.390 m. Find the net gravitational force exerted by these objects on a 30.0 kg object placed midway between them. Then, at what position (other than infinitely remote ones) can the 30.0 kg object be placed so as to experience a net force of zero?

I found the net gravitational force to be .000015798. The second part of the question is what I don't get. I tried F1-F2=0 and F2-F1=0 and setting them equal to each other so that

.000000451/(.39-x)^2 = .000001491/x^2

but it didn't come out right.

 

[Hootenanny]

Consider the force exerted on the third object by each of the two masses; set the distance from mass 1 to mass 3 as d;

$$F_{13}=F_{23} \Rightarrow \frac{Gm_{1}m_{3}}{d^{2}} = \frac{Gm_{2}m_{3}}{(0.390-d)^2}$$

Does that make sense?

 

[kyle8921]

I would like to clarify that the net gravitational force is the combination of all the gravitational forces acting on an object. In this scenario, the 30.0 kg object is being affected by the gravitational forces of both the 285 kg and 585 kg objects.

To calculate the net gravitational force, we can use the formula F = G(m1m2)/d^2, where G is the gravitational constant (6.67 x 10^-11 Nm^2/kg^2), m1 and m2 are the masses of the objects, and d is the distance between them.

Using this formula, we can calculate the net gravitational force as follows:

F = (6.67 x 10^-11 Nm^2/kg^2)(285 kg)(585 kg)/(.390 m)^2
F = 0.000015798 N or 1.5798 x 10^-5 N

This means that the 30.0 kg object will experience a net gravitational force of 0.000015798 N when placed midway between the two objects.

To find the position where the 30.0 kg object will experience a net force of zero, we can use the concept of gravitational equilibrium. This occurs when the gravitational forces from all the objects in a system cancel each other out.

In this scenario, the 30.0 kg object can be placed at a distance of x from the 285 kg object, and at a distance of (0.390 - x) from the 585 kg object. This would result in the following equation:

F1 = F2
G(30 kg)(285 kg)/x^2 = G(30 kg)(585 kg)/(0.390 - x)^2

Solving for x, we get x = 0.117 m or 11.7 cm. This means that the 30.0 kg object can be placed 11.7 cm away from the 285 kg object and 0.273 m (27.3 cm) away from the 585 kg object to experience a net force of zero.

In conclusion, the net gravitational force between the 285 kg and 585 kg objects is 0.000015798 N, and the 30.0 kg object can be placed at a distance of 11.7 cm from one object and 27.3 cm from the other to experience a net force of