Question

If the mass of a planet is eight times the mass of the earth and its radius is twice the radius of the earth, what will be the escape velocity of that planet?

Answer 1

escape velocity of earth 11.2$km/\sec$. The escape velocity depends only on the mass and size of the object from which something is trying to escape. The escape velocity from the Earth is the same for a pebble as it would be for the Space Shuttle.

Complete step by step answer:
let the mass of earth be ${M_{earth}}$ and the mass of planet be${M_{planet}}$.
According to the question we know that ${M_{planet}} = 8{M_{earth}}$
Let radius of earth be ${r_{earth}}$and radius of the planet be ${r_{planet}}$
According to the question we know that ${r_{planet}} = 2{r_{earth}}$
Equation to calculate the escape velocity of a planet. ${v_{escape}} = \sqrt {\dfrac{{2G{M_{PLANET}}}}{{{R_{planet}}}}}$
Where G is gravitational G = Newton's universal constant of gravity$6.67 \times {10^{ - 11}}N/{m^2}/k{g^3}$
M= mass of the planet; R = radius of the planet.
So to find the escape velocity for our planet.
${v_{escape}} = \sqrt {\dfrac{{2G8{M_{earth}}}}{{2{R_{earth}}}}}$
${v_{escape}} = (\sqrt {\dfrac{8}{2}} ) \times \sqrt {\dfrac{{2G{M_{earth}}}}{{{R_{earth}}}}}$
We already know that the escape velocity of earth is 11.2km/sec.
${v_{escape}} = \sqrt {\dfrac{8}{2}} \times 11.2$
$\therefore {v_{escape}}= 22.4km/\sec$

The escape velocity of our new planet will be 22.4 km/sec.

Note: Other method: We can solve this question by equating the ratio of planet earth and our new planet.
So to find we know
${v_{escape}} = \sqrt {2gR}$
If mass of the planet is 8 times that of earth
$\dfrac{{{v_{planet}}}}{{{v_{earth}}}} = \sqrt {\dfrac{{{M_{planet}}}}{{{M_{earth}}}} \times \Rightarrow\dfrac{{{\operatorname{R} _{earth}}}}{{{R_{planet}}}}}$
$\Rightarrow\dfrac{{{v_{planet}}}}{{{v_{earth}}}} = \sqrt {\dfrac{{8{M_{earth}}}}{{{M_{earth}}}} \times \Rightarrow\dfrac{{{R_{earth}}}}{{2{R_{earth}}}}} \\\$
$\Rightarrow\dfrac{{{v_{planet}}}}{{{v_{earth}}}} = 2$
$\Rightarrow{v_{planet}} = 2{v_{earth}}$
$\Rightarrow {v_{planet}} = 2 \times 11.2km/\sec$
$\therefore{v_{planet}} = 22.4km/\sec \\\$