Question 65: Gravitation

Question

The escape velocity from a spherical planet $A$ is $10 \mathrm{~km} / \mathrm{s}$. The escape velocity from another planet $B$ whose density and radius are $10 \%$ of those of planet $A$, is $\_\_\_\_$ $\mathrm{m} / \mathrm{s}$.

Accepted Answer

Escape velocity ( $\mathrm{v}_{\mathrm{e}}$ ) is the minimum initial velocity an object needs to completely escape a planet's gravitational field, meaning it reaches an infinite distance with zero remaining kinetic energy.
Let an object of mass $m$ be on the surface of a planet with mass $M$ and radius $R$.
Initial Energy at the surface:
The object has kinetic energy $\left(\frac{1}{2} \mathrm{mv}_{\mathrm{e}}^2ight)$ and gravitational potential energy $\left(-\frac{\mathrm{GMm}}{\mathrm{R}}ight)$.
Total Initial Energy $=\frac{1}{2} \mathrm{mv}_{\mathrm{e}}^2-\frac{\mathrm{GMm}}{\mathrm{R}}$
Final Energy at infinity:
The object just escapes, meaning it arrives with zero velocity (zero kinetic energy) and at an infinite distance, potential energy is also zero.
Total Final Energy $=0$
By the law of conservation of energy, Initial Energy = Final Energy:
$$ \frac{1}{2} m v_e^2-\frac{G M m}{R}=0 $$
$$\Rightarrow $$ $\frac{1}{2} \mathrm{mv}_{\mathrm{e}}^2=\frac{\mathrm{GMm}}{\mathrm{R}}$
$$ \mathrm{v}_{\mathrm{e}}=\sqrt{\frac{2 \mathrm{GM}}{\mathrm{R}}} $$
For a spherical planet, Volume (V) is $\frac{4}{3} \pi R^3$.
$$ \operatorname{Mass}(M)=\operatorname{Volume}(V) 	imes 	ext { Density }(ho) $$
$$ \mathrm{M}=\frac{4}{3} \pi \mathrm{R}^3 ho $$
Substituting this into escape velocity formula :
$$ \mathrm{v}_{\mathrm{e}}=\sqrt{\frac{2 \mathrm{G}}{\mathrm{R}} \cdot\left(\frac{4}{3} \pi \mathrm{R}^3 hoight)} $$
$$\Rightarrow $$ $\mathrm{v}_{\mathrm{e}}=\sqrt{\frac{8}{3} \pi \mathrm{GR}^2 ho}$
$$\Rightarrow $$ $\mathrm{v}_{\mathrm{e}}=\mathrm{R} \sqrt{\frac{8}{3} \pi \mathrm{G} ho}$
$$ \Rightarrow \mathrm{v}_{\mathrm{e}} \propto \mathrm{R} \sqrt{ho} $$
Escape velocity of planet $\mathrm{A}, \mathrm{v}_{\mathrm{eA}}=10 \mathrm{~km} / \mathrm{s}=10000 \mathrm{~m} / \mathrm{s}$
Radius of planet $B$ is $10 \%$ of radius of planet $A, R_B=10 \%$ of $R_A=0.1 R_A$
Density of planet B is $10 \%$ of that of planet A , $ho_{\mathrm{B}}=10 \%$ of $ho_{\mathrm{A}}=0.1 ho_{\mathrm{A}}$ Using proportionality relationship,
$$ \frac{\mathrm{v}_{\mathrm{eB}}}{\mathrm{v}_{\mathrm{eA}}}=\frac{\mathrm{R}_{\mathrm{B}} \sqrt{ho_{\mathrm{B}}}}{\mathrm{R}_{\mathrm{A}} \sqrt{ho_{\mathrm{A}}}} $$
$$\Rightarrow $$ $\frac{\mathrm{v}_{\mathrm{eB}}}{\mathrm{v}_{\mathrm{eA}}}=\frac{0.1 \mathrm{R}_{\mathrm{A}} \sqrt{0.1 ho_{\mathrm{A}}}}{\mathrm{R}_{\mathrm{A}} \sqrt{ho_{\mathrm{A}}}}$
$$\Rightarrow $$ $\frac{\mathrm{v}_{\mathrm{eB}}}{\mathrm{v}_{\mathrm{eA}}}=0.1 \sqrt{0.1}$
$$\Rightarrow $$ $\frac{\mathrm{v}_{\mathrm{eB}}}{\mathrm{v}_{\mathrm{eA}}}=\frac{1}{10} 	imes \frac{1}{\sqrt{10}}=\frac{1}{10 \sqrt{10}}$
$$\Rightarrow $$ $\mathrm{v}_{\mathrm{eB}}=10000 	imes\left(\frac{1}{10 \sqrt{10}}ight)$
$$ \mathrm{v}_{\mathrm{eB}}=\frac{1000}{\sqrt{10}}=100 \sqrt{10} \mathrm{~m} / \mathrm{s} $$
Therefore, the escape velocity of another planet B is $100 \sqrt{10} \mathrm{~m} / \mathrm{s}$.
Hence, the correct option is (D).

Answer

1000

Answer

$200 \sqrt{5}$

Answer

$1000 \sqrt{2}$

Answer

$100 \sqrt{10}$

Gravitation

The escape velocity from a spherical planet A is 10 \mathrm{~km} / \mathrm{s}. The escape velocity from another planet B whose density and radius are 10 \% of those of planet A, is \_\_\_\_ \mathrm{m} / \mathrm{s}.

Choose the correct answer: