Question

(a) What is the escape speed on a spherical asteroid whose radius is $500{\text{ km}}$ and whose gravitational acceleration at the surface is $3.0{\text{ m}}{{\text{s}}^{ - 2}}$.
(b) How far from the surface will a particle go if it leaves the asteroid’s surface with a radial speed of $1000{\text{ m}}{{\text{s}}^{ - 1}}$.
(c) With what speed will an object hit the asteroid’s surface if it is dropped from $1000{\text{ km}}$ above the surface.

Answer 1

Hint: We are first asked to find the escape velocity for which we need to express the formula for escape velocity in terms of gravitational acceleration. Then we are asked to find the height up to which a particle goes with a certain radial speed. Then we need to find the speed of the particle which is dropped from a height above the surface. We need to use the law of conservation of energy to solve the problem.

Complete step by step answer:
(a) The given values are
$r = 500{\text{ km}}$
$\Rightarrow a = 3.0{\text{ m}}{{\text{s}}^{ - 2}}$
$\Rightarrow v = 1000{\text{ m}}{{\text{s}}^{ - 1}}$
$\Rightarrow h = 1000{\text{ km}}$
We know that the formula for gravitational acceleration is given as
${a_g} = \dfrac{{GM}}{{{r^2}}}$
And the formula for the escape velocity is given by
$v = \sqrt {\dfrac{{2GM}}{r}} $
Substituting for gravitational acceleration in the above formula we get
$ \Rightarrow v = \sqrt {\dfrac{{2{a_g}{r^2}}}{r}} $
$ \Rightarrow v = \sqrt {2{a_g}r} $
Substituting the given values for the radius and gravitational acceleration.
$ \Rightarrow v = \sqrt {2(3.0)(500 \times {{10}^3}} )$
$\therefore v = 1.732 \times {{10}^3}{\text{ m}}{{\text{s}}^{ - 1}}$

(b) For finding the distance from the asteroid’s surface the particle to the height the particle would go, let the initial potential energy of the particle be ${U_i}$ and the initial kinetic energy of the particle be ${K_i}$ and let the required distance be $h$
We know ${U_i} = - \dfrac{{GMm}}{r}$
${K_i} = \dfrac{1}{2}m{v^2}$
Where $M$ is the mass of the asteroid and $m$ is the mass of the particle.
And when it reaches the distance its final potential energy will be ${U_f}$ and the final kinetic energy will be ${K_f}$ and it will be given as
${U_f} = - \dfrac{{GMm}}{{r + h}}$
$\Rightarrow{K_f} = 0$
Using the law of conservation of energy we get
${U_i} + {K_i} = {U_f} + {K_f}$
Substituting the values we get
$ - \dfrac{{GMm}}{r} + \dfrac{1}{2}m{v^2} = - \dfrac{{GMm}}{{r + h}} + 0$
$\Rightarrow - {a_g} + \dfrac{1}{2}{v^2} = - \dfrac{{{a_g}{r^2}}}{{r + h}}$
Rearranging the equation we get
$h = \dfrac{{2{a_g}{r^2}}}{{2{a_g}{r^2} - {v^2}}} - r$
Substituting the values we get
$h = \dfrac{{2(3.0)(500 \times {{10}^3})}}{{2(3.0)(500 \times {{10}^3}) - {{(1000)}^2}}} - (500 \times {10^3})$
$\therefore h = 2.5 \times {{10}^5}{\text{ m}}$

(c) For the speed of an object that will hit the asteroid if it is dropped above the surface. The distance from which the object is dropped will result in potential energy along with zero kinetic energy. Therefore we have,
${U_i} = - \dfrac{{GMm}}{{r + h}}$ and ${K_i} = 0$
And the final potential energy and the kinetic energy will be
${U_f} = - \dfrac{{GMm}}{r}$ and ${K_f} = \dfrac{1}{2}m{v^2}$
Again using the law of conservation of energy we get
$ - \dfrac{{GMm}}{{r + h}} + 0 = - \dfrac{{GMm}}{r} + \dfrac{1}{2}m{v^2}$
$\Rightarrow - \dfrac{{2{a_g}{r^2}}}{{r + h}} = - {a_g}r + \dfrac{1}{2}{v^2}$
Rearranging the equation we get
$v = \sqrt {2{a_g}r - \dfrac{{2{a_g}{r^2}}}{{r + h}}} $
$\Rightarrow v = \sqrt {2(3.0)(500 \times {{10}^3}) - \dfrac{{2(3.0)(500 \times {{10}^3})}}{{(500 \times {{10}^3}) + (1000 \times {{10}^3})}}} $
$\therefore v = 1.4 \times {{10}^3}{\text{ m}}{{\text{s}}^{ - 1}}$

Note: Escape velocity is the minimum speed needed for an object to escape from the gravitational influence of a massive body to eventually reach an infinite distance from it. We used the law of conservation of energy which states that the energy can neither be created nor be destroyed, it can only be converted into one form from another.