Question

In deep space, sphere Aof mass $\mathbf{20}\mathbf{}\mathit{k}\mathit{g}$is located at the origin of an x-axis and sphere Bof mass$\mathbf{10}\mathbf{}\mathit{k}\mathit{g}$ is located on the axis at$\mathbf{x}\mathbf{}\mathbf{=}\mathbf{0}\mathbf{.}\mathbf{80}\mathbf{}\mathit{m}$ . Sphere Bis released from rest while sphere Ais held at the origin. (a) What is the gravitational potential energy of the two-sphere system just as Bis released? (b) What is the kinetic energy of Bwhen it has moved $\mathbf{0}\mathbf{.}\mathbf{20}\mathbf{}\mathit{m}$toward A?

Short answer

a. The gravitational potential energy of the two-sphere system just as B is released is$-1.7\times {10}^{-8}\text{J}$

b. The kinetic energy of B when it has moved $0.20m$towards A is $5.6\times {10}^{-9}\text{J}$.

Step-by-step solution

Step 1: Given

Sphere A,${\text{M}}_{\text{A}}=20\text{kg}$at origin$\text{x}=0\text{m}$and

Sphere B, ${\text{M}}_{\text{B}}=10kg$ at $\text{x}=0.80\text{m}$

Determining the concept

Using the formula of gravitational potential energy and the principle of conservation of energy, find thegravitational potential energy of the two-sphere system just as B is released and the kinetic energy of B when it has moved towards Arespectively. According to the law ofconservation of energy, energy can neither be created nor be destroyed.

Formulae are as follows:

$$\begin{gathered} {\mathit{U}}_{\mathbf{i}}\mathbf{+}{\mathit{K}}_{\mathbf{i}}\mathbf{=}{\mathit{U}}_{\mathbf{f}}\mathbf{+}{\mathit{K}}_{\mathbf{f}} \\ \mathit{U}\mathbf{=}\frac{\mathbf{G}{\mathbf{M}}_{\mathbf{A}}{\mathbf{M}}_{\mathbf{B}}}{\mathbf{R}} \\ \mathit{K}\mathbf{=}\frac{\mathbf{1}}{\mathbf{2}}\mathit{m}{\mathit{v}}^{\mathbf{2}} \end{gathered}$$

where, ${\mathit{M}}_{\mathbf{A}}\mathbf{,}\mathbf{ }\mathbf{ }{\mathit{M}}_{\mathbf{B}}$, m are masses, R is the radius, v is velocity, G is gravitational constant, K is kinetic energy and U is potential energy.

(a) Determining the gravitational potential energy of the two-sphere system just as   is released

Now,

$U_i=-\frac{G{M}_A{M}_B}{r_i}$

As

$\begin{array}{rcl}{M}_A& =& 20\text{kg}, M_B=10\text{kg},and r_i=0.80\text{m}\\ U_i& =& -\frac{\left({\text{6.67×10}}^{\text{- 11}}\frac{{\text{m}}^{\text{3}}}{{\text{s}}^{\text{2}}·\text{kg}}\right)\left(\text{20}\text{kg}\right)\left(\text{10}\text{kg}\right)}{0.80m}\\ & =& -1.7\times {10}^{-8}J\\ & & \end{array}$

Hence, the gravitational potential energy of the two-sphere system just as B is released is $-1.7\times {10}^{-8}J$.

(b) Determining the kinetic energy of  B when it has moved  0.20m towards  A

Now,

$U_i+K_i=U_f+K_f$

As

$\begin{array}{rcl}Ui& =& -1.7\times {10}^{-8}J,and{K}_i=0\\ r_i& =& 0.\text{80}\text{m} -\text{0.20}\text{m}\\ & =& \text{0.60}\text{m}\\ & & \\ & & \end{array}$

$\begin{array}{rcl}-1.7\times {10}^{-8}J& =& K-\frac{\left({\text{6.67×10}}^{\text{- 11}}\frac{{\text{m}}^{\text{3}}}{{\text{s}}^{\text{2}}·\text{kg}}\right)\left(\text{20}\text{kg}\right)\left(\text{10}\text{kg}\right)}{0.60m}\\ -1.7\times {10}^{-8}J& =& K-2.22\times {10}^{-8}J\\ K& =& 5.6\times {10}^{-9}J\end{array}$

Hence, the kinetic energy of when it has moved $0.20m$towards A is $5.6\times {10}^{-9}\text{J.}$

Therefore, using the formula for gravitational potential energy and the law of conservation of energy, kinetic energy can be found.