Q15E - University Physics with Modern Physics

Question

Inside a starship at rest on the earth, a ball rolls off the top of a horizontal table and lands a distance D from the foot of the table. This starship now lands on the unexplored Planet X. The commander, Captain Curious, rolls the same ball off the same table with the same initial speed as on earth and finds that it lands a distance 2.76D from the foot of the table. What is the acceleration due to gravity on Planet X?

Step-by-Step Solution

Answer

The acceleration due to gravity on planet X is $1.283\hspace{0.33em}\mathrm{m}/{\mathrm{s}}^2$.

Step 1: Identification of given data:

The given data can be listed below.

 

Acceleration due to gravity,  $\mathrm{g}=9.8\hspace{0.33em}\mathrm{m}/{\mathrm{s}}^2$.

 

The distance of ball from foot of the table is ${\mathrm{R}}_{\mathrm{E}}=\mathrm{D}$.

 

The distance of ball from foot of table on earth is ${\mathrm{R}}_{\mathrm{x}}=2.76\mathrm{D}$. 

Step 2: Concept/Significance of gravitational acceleration:

The acceleration owing to gravity is the acceleration of a body caused by the influence of gravity's pull alone and is typically indicated by the letter  . Each celestial body has a unique value for this quantity.

Step 3: Determination of the acceleration due to gravity on Planet X:

The horizontal range of the ball is given by,

$\mathrm{R}={\mathrm{v}}_0\sqrt{\frac{2\mathrm{h}}{\mathrm{g}}}$  

 

Here, h is height, ${\mathrm{v}}_0$ is the velocity, and g is the acceleration due to gravity.

 

The ratio of range of planet X and Earth is given by,

$\frac{{\mathrm{R}}_{\mathrm{E}}}{{\mathrm{R}}_{\mathrm{x}}}=\sqrt{\frac{{\mathrm{g}}_{\mathrm{x}}}{{\mathrm{g}}_{\mathrm{E}}}}$ 

 

Substitute all the values in the above,

$$\begin{gathered} \frac{\mathrm{D}}{2.76\mathrm{D}}=\sqrt{\frac{{\mathrm{g}}_{\mathrm{x}}}{9.8\mathrm{m}/{\mathrm{s}}^2}} \\ 0.131=\frac{{\mathrm{g}}_{\mathrm{x}}}{9.8\mathrm{m}/{\mathrm{s}}^2} \\ {\mathrm{g}}_{\mathrm{x}}=\left(0.131\right)\left(9.8\mathrm{m}/{\mathrm{s}}^2\right) \\ =1.283\mathrm{m}/{\mathrm{s}}^2 \end{gathered}$$ 

 Thus, the acceleration due to gravity on planet X is $1.283\mathrm{m}/{\mathrm{s}}^2$ .