Question

An astronaut circling the earth at an altitude of $400km$is horrified to discover that a cloud of space debris is moving in the exact same orbit as his spacecraft, but in the opposite direction. The astronaut detects the debris when it is $25km$away. How much time does he have to fire his rockets and change orbits?

Short answer

The time required to fire his rockets and change orbits is$1.6sec.$

Step-by-step solution

Given information

height of the astronaut $h=400km=4.0\times {10}^{5}m$

Gravitational constant $G=6.67\times {10}^{-11}{m}^3/kg{s}^2$

Radius of earth role="math" localid="1649933256327" $R=6.37\times {10}^{6}m$

distance between debris and astronaut$d=25000m$

Explanation

One of the essential concepts to solve this problem is the utilization of the equations of centripetal and gravitational force.

From them it will be possible to find the speed of the body with which the estimated time can be calculated through the kinematic equations of motion. At the same time for the calculation of this speed it is necessary to clarify that this will remain twice the ship, because as we know by relativity, when moving in the same magnitude but in the opposite direction, with respect to the ship the debris will be double speed.

By equilibrium the centrifugal force and the gravitational force are equal therefore

$$\begin{gathered} F_c=F_g \\ \frac{m{v^2}_{orbit}}{r}=\frac{GMm}{r^2} \end{gathered}$$

Where

$m$is mass spacecraft

$v$ is velocity

$G$is Gravitational Universal Constant

$M$is Mass of earth

Radius of earth and orbit $r=R+h$

Re-arrange to find the velocity

$$\begin{gathered} \frac{m{v^2}_{orbit}}{r}=\frac{GMm}{r^2} \\ {v}_{orbit}=\sqrt{\frac{GM}{R+h}} \end{gathered}$$

Replacing with our values we have

role="math" localid="1649934164508" $$\begin{gathered} v_{orbit}=\sqrt{\frac{6.67\times {10}^{-11}\times 5.98\times {10}^{24}}{6.37\times {10}^6+0.4\times {10}^6}} \\ {v}_{orbit}=7852m/s \end{gathered}$$

From the cinematic equations of motion we have to

$t=\frac{d}{2v_{orbit}}$Remember that the speed is double for the counter-direction of the trajectories.

Replacing

role="math" localid="1649934211102" $t=\frac{25000}{2\times 7852}=1.6sec$

Therefore the time required isrole="math" localid="1649934227954" $1.6sec.$