Question

Figure 13-46a shows a particleAthat can be movedalong ay-axis from an infinite distance to the origin. That origin liesat the midpoint between particlesBandC, which have identical masses, and theyaxis is a perpendicular bisector between them.DistanceDis$\mathbf{0}\mathbf{.}\mathbf{3057}\mathbf{}\mathbf{\text{m}}$. Figure 13-46b shows the potential energyUof the three-particle system as a function of the position of particleAalong theyaxis. The curve actually extends rightward and approaches an asymptote of$\mathbf{-}\mathbf{2}\mathbf{.}\mathbf{7}\mathbf{\times }{\mathbf{10}}^{\mathbf{11}}\mathbf{\text{J}}$as$\to \infty$. What are themasses of (a) particlesBandCand (b) particleA?

Short answer

1. Masses of particlesB and C are $0.50\text{kg}$.
2. Mass of particle A is $1.5\text{kg}$.

Step-by-step solution

Step 1: Given

Distance,D is$\text{0.3057m}$

Determining the concept

Using the formula for gravitational potential energy, find the masses of particles B, C, and A for an infinitely large distance y and y = 0.

The formula is as follows:

$\mathit{U}\mathbf{=}\mathbf{-}\frac{\mathbf{G}\mathbf{M}\mathbf{m}}{\mathbf{R}}$

where, m, and M are masses, R is the radius, G is gravitational constant and U is potential energy.

(a) Determining themasses of particles B and C

Now,

$U=-\frac{\mathrm{GMm}}{\mathrm{R}}$

Applying the Pythagorean Theorem to a graph gives,

$U=-\left(\frac{G{M}^2}{2D}+\frac{2GmM}{\sqrt{y^2+D^2}}\right)$

where M is the mass of particles B and C is the mass of a particle A .

For an infinitely large term,

$$\begin{gathered} \frac{2GmM}{\sqrt{y^2+D^2}}=0 \\ \text{D}=0.3057\text{m} \\ U=-\left(\frac{G{M}^2}{2D}\right) \\ M=0.50kg \end{gathered}$$

Therefore, the masses of a particle $BandC$ are $0.50kg$.