Q47E - University Physics with Modern Physics

Question

A 15-kg rock is dropped from rest on the earth and reaches the ground in 1.75 s. When it is dropped from the same height on Saturn’s satellite Enceladus, the rock reaches the ground in 18.6 s. What is the acceleration due to gravity on Enceladus?

Step-by-Step Solution

Answer

Acceleration due to gravity on Enceladus =$0.0867 \mathrm{m}/{\mathrm{s}}^2$  

Step 1: Identification of given values

Rock dropped Time from earth (t₁) = 1.75 s

Mass = 15 kg

Rock dropped Time from Saturn’s satellite Enceladus (t₂) = 18.6 s

Step 2: calculation for acceleration due to gravity

Let the height of drop be h,

For a body, dropped from rest, this height is given by

 $\mathrm{h}=\frac{1}{2}{\mathrm{g}}_{\mathrm{e}}{\mathrm{t}}^2$

 $$\begin{gathered} \mathrm{for} \mathrm{earth} {\mathrm{g}}_{\mathrm{e}} =9.8\mathrm{m}/{\mathrm{s}}^2 \\ \mathrm{h}=\frac{1}{2}\times 9.8\times 1.{75}^2 \\ =15 \mathrm{m} \end{gathered}$$

For a body, dropped Saturn’s satellite Enceladus,

$$\begin{gathered} h=\frac{1}{2}{g}_s{t}_2^2 \\ {g}_s=\frac{2h}{t_2^2} \end{gathered}$$ 

Where ${\mathrm{g}}_{\mathrm{s}}$is gravity of Saturn’s satellite Enceladus

Put h=15m and${\mathrm{t}}_2=18.6_{\mathrm{s}}$

$$\begin{gathered} {\mathrm{g}}_{\mathrm{s}}=\frac{2\times 15}{18.6^2} \\ =0.0867 \mathrm{m}/{\mathrm{s}}^2 \end{gathered}$$