Question

In Problems 64 through 66 you are given the equation(s) used to
solve a problem. For each of these, you are to
a. Write a realistic problem for which this is the correct equation(s).
b. Draw a pictorial representation.
c. Finish the solution of the problem.

$66.\frac{1}{2}(100\mathrm{kg})v_2^2-\frac{\left(6.67\times {10}^{-11}{\mathrm{Nm}}^2/{\mathrm{kg}}^2\right)\left(7.36\times {10}^{22}\mathrm{kg}\right)(100\mathrm{kg})}{1.74\times {10}^6\mathrm{m}}=0-\frac{\left(6.67\times {10}^{-11}{\mathrm{Nm}}^2/{\mathrm{kg}}^2\right)\left(7.36\times {10}^{22}\mathrm{kg}\right)(100\mathrm{kg})}{3.48\times {10}^6\mathrm{m}}$

Short answer

a) A 100 kg object is at the moon surface, it is required to be shift at a distance of 3.48 x 10⁶ m from the moon surface. Find the velocity of the object on the moon surface so that it can reach the given distance.

b) Pictorial representation of the problem is as below

c)

Step-by-step solution

Part(a)Step1: Given information

The given equation is

$\frac{1}{2}(100\mathrm{kg})v_2^2-\frac{\left(6.67\times {10}^{-11}{\mathrm{Nm}}^2/{\mathrm{kg}}^2\right)\left(7.36\times {10}^{22}\mathrm{kg}\right)(100\mathrm{kg})}{1.74\times {10}^6\mathrm{m}}=0-\frac{\left(6.67\times {10}^{-11}{\mathrm{Nm}}^2/{\mathrm{kg}}^2\right)\left(7.36\times {10}^{22}\mathrm{kg}\right)(100\mathrm{kg})}{3.48\times {10}^6\mathrm{m}}$

Part(a) Step2: Explanation

The given equation appears to be comparison of energy of the same object at two different places

We can frame the problem as

A 100 kg object is at the moon surface, it is required to be shift at a distance of 3.48 x 10⁶ m from the moon surface. Find the velocity of the object on the moon surface so that it can reach the given distance.

Part(b)Step1: Given information

The given equation is

$$\begin{gathered} \frac{1}{2}(100\mathrm{kg})v_2^2-\frac{\left(6.67\times {10}^{-11}{\mathrm{Nm}}^2/{\mathrm{kg}}^2\right)\left(7.36\times {10}^{22}\mathrm{kg}\right)(100\mathrm{kg})}{1.74\times {10}^6\mathrm{m}} \\ =0-\frac{\left(6.67\times {10}^{-11}{\mathrm{Nm}}^2/{\mathrm{kg}}^2\right)\left(7.36\times {10}^{22}\mathrm{kg}\right)(100\mathrm{kg})}{3.48\times {10}^6\mathrm{m}} \end{gathered}$$

Part(b) Step2: Explanation

The pictorial representation of the problem is as below

Part(c)Step1 : Given information

The given expression is

$$\begin{gathered} \frac{1}{2}(100\mathrm{kg})v_2^2-\frac{\left(6.67\times {10}^{-11}{\mathrm{Nm}}^2/{\mathrm{kg}}^2\right)\left(7.36\times {10}^{22}\mathrm{kg}\right)(100\mathrm{kg})}{1.74\times {10}^6\mathrm{m}} \\ =0-\frac{\left(6.67\times {10}^{-11}{\mathrm{Nm}}^2/{\mathrm{kg}}^2\right)\left(7.36\times {10}^{22}\mathrm{kg}\right)(100\mathrm{kg})}{3.48\times {10}^6\mathrm{m}} \end{gathered}$$

Part(c): Step2: Explanation

Solve the given equation to find the velocity

$$\begin{gathered} \frac{1}{2}(100\mathrm{kg})v_2^2-\frac{\left(6.67\times {10}^{-11}{\mathrm{Nm}}^2/{\mathrm{kg}}^2\right)\left(7.36\times {10}^{22}\mathrm{kg}\right)(100\mathrm{kg})}{1.74\times {10}^6\mathrm{m}}=0-\frac{\left(6.67\times {10}^{-11}{\mathrm{Nm}}^2/{\mathrm{kg}}^2\right)\left(7.36\times {10}^{22}\mathrm{kg}\right)(100\mathrm{kg})}{3.48\times {10}^6\mathrm{m}} \\ 50kg{v}_2^2=(6.67\times {10}^{-11}{\mathrm{Nm}}^2/{\mathrm{kg}}^2)\times (7.36\times {10}^{22}kg)\times \left(100kg\right)\left(\frac{1}{1.74\times {10}^{6}m}-\frac{1}{3.48\times {10}^{6}m}\right) \\ v=1680m/s \end{gathered}$$