Question

Mr. Bean can jump to a height of 1.5 m on earth. What is the height he may be able to jump on another planet whose density and radius are, respectively, one-quarter and one-third of earth?
1) 1.5m
2) 15m
3) 18m
4) 28m

Answer 1

Hint:Every object in the universe attracts every other object with a force which is directly proportional to the product of their masses and inversely proportional to the square of the distance between them. Apply the formula for both planets and equate it with the formula for height $h = \dfrac{{{u^2}}}{{2g}}$.

Complete step by step solution:
Apply the formula for gravitational force:
$F = G\dfrac{{mM}}{{{r^2}}}$ ;
Now, we know that F = mg. Put it in the above formula
$ \Rightarrow mg = G\dfrac{{mM}}{{{r^2}}}$;
Cancel out the common:
$ \Rightarrow g = \dfrac{{GM}}{{{r^2}}}$;
Now for Earth$M = D \times V$:
 $g = \dfrac{{G\left( {D \times V} \right)}}{{{r^2}}}$;
The volume of the sphere is $V = \dfrac{4}{3}\pi {r^3}$.
$g = \dfrac{{G\left( {\rho \times \left( {4/3} \right)\pi {r^3}} \right)}}{{{r^2}}}$;
$g = \dfrac{4}{3}\left( {G\pi } \right)\rho r$;
Now, Similarly for another planet:
It is given that ${\rho _p} = \dfrac{\rho }{4}$ and ${R_p} = \dfrac{r}{3}$ ;
$g' = \dfrac{4}{3}\left( {G\pi } \right){\rho _p}{R_p}$
Put the value of density and radius of another planet in the above equation:
$ \Rightarrow g' = \dfrac{4}{3}\left( {G\pi } \right)\left( {\dfrac{\rho }{4}} \right)\dfrac{r}{3}$;
$ \Rightarrow g' = \dfrac{1}{{12}}\left( {\dfrac{4}{3}G\pi } \right)\rho r;$
Now, we have found out that $g' = \dfrac{4}{3}\left( {G\pi } \right)\rho r$:
$ \Rightarrow g' = \dfrac{1}{{12}}g;$
$ \Rightarrow \dfrac{g}{{g'}} = 12;$
We know the formula for height:
${u^2} = 2gh$;
 $ \Rightarrow \dfrac{{{u^2}}}{{2g}} = h$;
Similarly, we can write the formula of height for another planet:
\[ \Rightarrow \dfrac{{{u^2}}}{{2g'}} = h'\];
Divide both the heights:
$\dfrac{{{u^2}}}{{2g}} \times \dfrac{{2g'}}{{{u^2}}} = \dfrac{h}{{h'}}$;
Do the necessary calculation:
\[ \Rightarrow \dfrac{{g'}}{g} = \dfrac{h}{{h'}}\];
Put the value $\dfrac{g}{{g'}} = 12;$in the above equation:
\[ \Rightarrow \dfrac{1}{{12}} = \dfrac{h}{{h'}}\];
We have been given the height as 1.5m
\[ \Rightarrow \dfrac{1}{{12}} = \dfrac{{1.5}}{{h'}}\]
\[ \Rightarrow h' = 12 \times 1.5 = 18m\];

Option “3” is correct. The height he may be able to jump on another planet whose density and radius are, respectively, one-quarter and one-third of earth is 18m

Note:Here, we first have to apply the gravitational force and find out the gravitational acceleration and then we have to find the gravitational acceleration by the same method for another planet. Put in the values of given in terms of density and radius after that apply the formula for height for both earth and the planet and equate the ratio of the gravitational acceleration and height.