What Is the Initial Acceleration of the Third Sphere in a Gravitational System?

[kai89]

Homework Statement

3 uniform spheres are located at the corners of an equilateral triangle. each side of the triangle has a length of 1.2 m. two of the spheres have a mass of 2.8 kg. the 3rd sphere (mass unknown) is released from rest. Considering only the gravitation forces the spheres have on each other, what is the magnitude of the initial acceleration of the 3rd sphere?

Homework Equations

F= G m1m2/r2

possibly 1 of the kinemetic equations

The Attempt at a Solution

I know that the force of the first 2 masses is 2.59x10-10 N. I am not sure how to finish the 2nd part of the problem. I know that the 3rd mass' initial velocity=0, but that's all. Can someone help me? I got a 51 on a open note open book quiz in this quiz last week, extremely pathetic. Thanks a lot!

 

[lanedance]

what does the force of the first two masses mean? a force acts on something...

consider the force form each mass (i=1,2) on the test mass say M

$$\textbf{F}_i = \frac{G m_i M}{r^2} \hat{\textbf{r}}$$
the bold r hat it is a vector pointing towards m_i

now the acceleration of Mass m is proportional to the net force
$$\textbf{a} = \frac{\sum_i \textbf{F}_i }{M}$$
note that this is vector addition of the forces

see anything that cancels...

 

[tomcue]

Based on the given information, we can use the equation for Newton's second law to solve for the initial acceleration of the 3rd sphere. This law states that the net force acting on an object is equal to its mass multiplied by its acceleration (F=ma).

In this case, the only force acting on the 3rd sphere is the gravitational force from the other two spheres. Therefore, we can set up the following equation:

F = m3a

Where F is the net gravitational force on the 3rd sphere and m3 is its mass. We can calculate the net gravitational force using the formula F=Gm1m2/r^2, where G is the gravitational constant, m1 and m2 are the masses of the first two spheres, and r is the distance between them.

Substituting the given values, we get:

F = (6.67x10^-11 Nm^2/kg^2)(2.8 kg)(2.8 kg)/(1.2 m)^2 = 2.59x10^-10 N

Now, we can plug this value into our equation for Newton's second law:

2.59x10^-10 N = m3a

Since we are solving for the initial acceleration (a), we can rearrange the equation to solve for a:

a = 2.59x10^-10 N/m3

We still need to find the mass of the 3rd sphere (m3) to get our final answer. To do this, we can use the fact that the total mass of the system is equal to the sum of the individual masses (m1 + m2 + m3). Since we know the masses of the first two spheres (2.8 kg each), we can rearrange this equation to solve for m3:

m3 = (total mass) - (m1 + m2) = m3 = (2.8 kg + 2.8 kg) - (2.8 kg + 2.8 kg) = 2.8 kg

Finally, we can substitute this value into our equation for acceleration and get:

a = 2.59x10^-10 N/2.8 kg = 9.25x10^-11 m/s^2

Therefore, the magnitude of the initial acceleration of the 3rd sphere is 9.25x10^-11 m/s^2.