Question

One way to attack a satellite in earth orbit is to launch a swarm of pellets in the same orbit as the satellite but in the opposite direction. Suppose a satellite in a circular orbit \[500{\text{km}}\] above Earth’s surface collides with a pallet having mass $4.0{\text{g}}$ . A) What is the kinetic energy of the pellet in the reference frame of the satellite just before the collision. B) What is the ratio of this kinetic energy to the kinetic energy of a $4.0{\text{g}}$ bullet from a modern army rifle with a muzzle speed $950{\text{m}}{{\text{s}}^{ - 1}}$.

Answer 1

Hint: We are given a pellet of certainty which is launched in the satellite’s orbit in the opposite direction to that of the satellite’s motion and we need to find the kinetic energy of the pellet about the satellite’s frame. We need first to calculate the velocity of the pellet for finding the kinetic energy.

Complete step by step answer:
$(a)$ Let the velocity of the pellets be $v$ now since, to attack the satellite, the pellets are launched in the opposite direction to that of the satellite’s motion therefore about the frame of the satellite, the velocity of the pellets will be twice the launch velocity. So the relative velocity between the pellets and the satellite is $2v$.

We know that the relative kinetics of a pellet escaping the earth’s gravitational force is derived from the escape velocity formula for a particle.Therefore we have
$K = 4\left( {\dfrac{{GMm}}{{2r}}} \right)$
Where $M$ is the mass of the earth, $m$ is the mass of the particle escaping which is the pellet, in this case, $G$ is gravitational constant and $r$ is the radius of the earth.
Hence by substituting the values we get
$K = \dfrac{{2(6.67 \times {{10}^{ - 11}})(5.98 \times {{10}^{24}})(0.0040)}}{{(6370 + 500) \times {{10}^3}}} \\
\therefore K= 4.6 \times {10^5}{\text{J}}$

$(b)$ In the second part, we need to find the ratio of the kinetic energy of the pellet to the kinetic energy of the bullet from a modern army rifle with a muzzle speed of $950{\text{m}}{{\text{s}}^{ - 1}}$
First, let us calculate the kinetic energy of the bullet
We know ${K_{{\text{bullet}}}} = \dfrac{1}{2}m{v^2}$
Substituting the value we get
${K_{{\text{bullet}}}} = \dfrac{1}{2} \times 0.0040 \times 950$
The required ratio
$\dfrac{K}{{{K_{{\text{bullet}}}}}} = \dfrac{{4.6 \times {{10}^5}{\text{J}}}}{{\dfrac{1}{2} \times 0.0040 \times 950}} \\
\therefore \dfrac{K}{{{K_{{\text{bullet}}}}}}= 2.6 \times {10^2}{\text{J}}$

Note: Note that before calculating the required values of kinetic energy we convert the CGS units to MKS units so that we get the value of kinetic energy in joules.This ensures smooth calculation. The gravitational constant used is the constant of proportionality in Newton’s law of universal gravitational.