The first known collision between space debris and a functioning satellite occurred in 1996: At an altitude of$\mathbf{\text{700km}}$, a year old French spy satellite was hit by a piece of an Ariane rocket. A stabilizing boom on the satellite was demolished, and the satellite was sent spinning out of control. Just before the collision and in kilometresper hour, what was the speed of the rocket piece relative to the satellite if both were in circular orbits and the collision was

(a) head-on and

(b) along perpendicular paths?

The altitude is $\text{700km}$.

Using the formula for orbital speed, findthe speed of the rocket piece relative to the satellite using the given altitude and mass of the object in circular orbit. The collision was head-on and along perpendicular paths.

Formula is as follow:

$\mathrm{v}=\sqrt{\frac{\mathrm{GM}}{\mathrm{R}}}$

where, v is velocity, G is gravitational constant, M is mass and R is radius.

Now,

$\mathrm{v}=\sqrt{\frac{\mathrm{GM}}{\mathrm{R}}}$

But the speed of an object in a circular orbit of mass,$\text{M}=5.98\times {10}^{24}\text{\hspace{0.17em}kg}$and radius,

$\begin{array}{c}\mathrm{R}=(\text{700\hspace{0.17em}m}+\text{6370\hspace{0.17em}}\mathrm{k}\text{m})\\ =\text{7070km}\\ =\text{7.07}\times {\text{10}}^{\text{6}}\text{\hspace{0.17em}m}\end{array}$

$\begin{array}{c}v=\sqrt{\frac{\left(\text{6.67}\times {\text{10}}^{\text{-11}}\frac{\text{N}\cdot {\text{m}}^{\text{2}}}{{\text{kg}}^{\text{2}}}\right)\left({\text{5.98×10}}^{\text{24}}\text{\hspace{0.17em}kg}\right)}{\text{7.07}\times {\text{10}}^{\text{6}}\text{\hspace{0.17em}m}}}\\ =7.51\times {10}^3\text{\hspace{0.17em}m/s}\\ =2.7\times {10}^4\text{\hspace{0.17em}km/h}\end{array}$

But, for head on collision, the speed is$\text{2v}$, because both the objects are moving at the same speed towards each other. Hence,

$\mathrm{v}=5.4\times {10}^4\text{\hspace{0.17em}km/h}$

Therefore, the speed of the rocket piece relative to the satellite if both were in circular orbits and the collision was head-on is $\text{v}=5.4\times {10}^4\text{\hspace{0.17em}km/h}$.

For perpendicular collision, the relative speed is given by Pythagorean Theorem,

$\mathrm{v}=\sqrt{{\mathrm{v}}^2+{\mathrm{v}}^2}$

$\begin{array}{c}\mathrm{v}=\sqrt{{\left(\text{2.7}\times {\text{10}}^{\text{4}}\frac{\text{km}}{\text{h}}\right)}^{\text{2}}\text{+}{\left(\text{2.7}\times {\text{10}}^{\text{4}}\frac{\text{km}}{\text{h}}\right)}^{\text{2}}}\\ =3.8\times {10}^4\mathrm{km}/\mathrm{h}\end{array}$

Hence, thespeed of the rocket piece relative to the satellite if both were in circular orbits and the collision was along perpendicular paths is${\text{3.8×10}}^{\text{4}}\text{km/h}$.

Therefore, using the formula for orbital velocity, and the concept of relative motion and Pythagoras Theorem, the relative speed can be found.