Objects with masses of 280 kg and a 580 kg are separated by 0.430 m.
(a)
Find the net gravitational force exerted by these objects on a 65.0 kg object placed midway between them.
magnitude	
.000028137
 
Correct: Your answer is correct.
 N
direction	
toward the 580 kg mass
 
Correct: Your answer is correct.
(b)
At what position (other than infinitely remote ones) can the 65.0 kg object be placed so as to experience a net force of zero?

### Part (b) Solution:

To find the position where the 65.0 kg object experiences a net gravitational force of zero, we must set the gravitational forces exerted by the two objects on the 65.0 kg mass equal in magnitude but opposite in direction.

Let the distance from the 65.0 kg mass to the 280 kg mass be \( x \), and the distance to the 580 kg mass be \( 0.43 - x \).

The gravitational force between two masses \( m_1 \) and \( m_2 \) separated by a distance \( r \) is given by Newton's law of gravitation:

\[
F = \frac{G \cdot m_1 \cdot m_2}{r^2}
\]

Where:
- \( G = 6.674 \times 10^{-11} \, \text{N} \cdot \text{m}^2/\text{kg}^2 \) (gravitational constant),
- \( m_1 = 280 \, \text{kg} \),
- \( m_2 = 580 \, \text{kg} \),
- \( m_3 = 65.0 \, \text{kg} \),
- \( r \) is the distance between the objects.

We will set the gravitational force exerted by the 280 kg mass equal to the gravitational force exerted by the 580 kg mass on the 65.0 kg object:

\[
\frac{G \cdot 280 \cdot 65}{x^2} = \frac{G \cdot 580 \cdot 65}{(0.43 - x)^2}
\]

Since \( G \) and the 65.0 kg mass cancel out from both sides, we are left with:

\[
\frac{280}{x^2} = \frac{580}{(0.43 - x)^2}
\]

### Cross-multiply:

\[
280 \cdot (0.43 - x)^2 = 580 \cdot x^2
\]

### Expanding both sides:

\[
280 \cdot (0.43^2 - 2 \cdot 0.43 \cdot x + x^2) = 580 \cdot x^2
\]

\[
280 \cdot (0.1849 - 0.86x + x^2) = 580 \cdot x^2
\]

### Simplify:

\[
51.772 - 240.8x + 280x^2 = 580x^2
\]

### Bring all terms to one side:

\[
280x^2 - 580x^2 - 240.8x + 51.772 = 0
\]

\[
-300x^2 - 240.8x + 51.772 = 0
\]

### Solving this quadratic equation using the quadratic formula:

The quadratic formula is:

\[
x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}
\]

Where \( a = -300 \), \( b = -240.8 \), and \( c = 51.772 \).

\[
x = \frac{-(-240.8) \pm \sqrt{(-240.8)^2 - 4 \cdot (-300) \cdot 51.772}}{2 \cdot (-300)}
\]

First, calculate the discriminant:

\[
\Delta = (-240.8)^2 - 4 \cdot (-300) \cdot 51.772
\]

\[
\Delta = 58048.64 + 62126.4
\]

\[
\Delta = 120175.04
\]

Now calculate \( x \):

\[
x = \frac{240.8 \pm \sqrt{120175.04}}{-600}
\]

\[
x = \frac{240.8 \pm 346.67}{-600}
\]

Now, calculate the two possible values for \( x \):

1. \( x_1 = \frac{240.8 + 346.67}{-600} = \frac{587.47}{-600} = -0.979 \, \text{m} \) (which is not physically valid because \( x \) must be positive).

2. \( x_2 = \frac{240.8 - 346.67}{-600} = \frac{-105.87}{-600} = 0.176 \, \text{m} \).

Thus, the position where the 65.0 kg object experiences zero net gravitational force is **0.176 m** from the 280 kg mass.

### Final Answer:
The 65.0 kg object should be placed **0.176 m from the 280 kg mass** (or equivalently, \( 0.43 - 0.176 = 0.254 \, \text{m} \) from the 580 kg mass).

---

Do you want details on any step or have more questions about the explanation?

### Related questions:
1. How do we derive the gravitational force equation from Newton's law of gravitation?
2. What assumptions are made when solving for equilibrium positions in gravitational problems?
3. How can this concept be extended to more than two masses?
4. What is the physical meaning of the quadratic formula in this context?
5. How does the gravitational constant \( G \) influence the force in different scenarios?

**Tip**: Always check units and physical constraints, like distances and directions, when solving for equilibrium points in gravitational problems.

Objects with masses of 280 kg and 580 kg are separated by 0.430 m. (a) Find the net gravitational force exerted by these objects on a 65.0 kg object placed midway between them. (b) At what position (other than infinitely remote ones) can the 65.0 kg object be placed so as to experience a net force of zero?

Discover how to calculate the gravitational force between two masses (280 kg and 580 kg) and a 65.0 kg object placed midway. Learn where to position the 65.0 kg object for zero net force using Newton's Law of Gravitation and quadratic equations. Suitable for students in Grades 10-12.

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