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Q6.6-48PE

Expert-verifiedFound in: Page 224

Book edition
1st Edition

Author(s)
Paul Peter Urone

Pages
1272 pages

ISBN
9781938168000

**Space debris left from old satellites and their launchers is becoming a hazard to other satellites**

**(a) Calculate the speed of a satellite in an orbit ** **900 km above Earth’s surface.**

**(b) Suppose a loose rivet is in an orbit of the same radius that intersects the satellite’s orbit at an angle of **${\mathbf{90}}{\mathbf{\xb0}}$** relative to Earth. What is the velocity of the rivet relative to the satellite just before striking it?**

**(c) Given** **the rivet is** **3.00 mm in size, how long will its collision with the satellite last?**

**(d) If its mass is** **0.500 g, what is the average force it exerts on the satellite? (e) How much energy in joules is generated by the collision? (The satellite’s velocity does not change appreciably, because its mass is much greater than the rivet’s.)**

- A satellite in orbit moves at a speed of ${\text{7.4\xd710}}^{\text{3}}\text{\hspace{0.33em}}\frac{\text{m}}{\text{s}}$.
- The velocity of the rivet w.r.t. the satellite is ${\text{1.05\xd710}}^{\text{4}}\text{\hspace{0.33em}}\frac{\text{m}}{\text{s}}$.
- The time in which collision will last is ${\text{2.86\xd710}}^{\text{-7}}\text{\hspace{0.33em}s}$.
- The average force that rivet exerts on the satellite is ${\text{1.83\xd710}}^{\text{7}}\text{\hspace{0.33em}N}$.
- The energy generated in the collision is ${\text{2.756\xd710}}^{\text{4}}\text{\hspace{0.33em}J}$.

Objects that orbit around other objects are classified as satellites. Moon orbits around the earth, and the earth orbits around the sun. Moon is the satellite of the earth, and the earth is a satellite of the sun.

There are satellites that are placed in orbits around planets. These are artificial satellites.

The gravitational force between the two objects is given by,

${F}_{g}=\frac{G{m}_{1}{m}_{2}}{{r}^{2}}$

Where *G* is the universal gravitation constant, *r* is the distance between the two objects, and ${\text{m}}_{\text{1}}{\text{,m}}_{\text{2}}$ are the masses of two bodies, respectively.

In angular motion, the centripetal acceleration is given,

$\begin{array}{c}{F}_{c}=m{a}_{c}\\ =m{\omega}^{2}r\end{array}$

As the satellite revolves around the earth, so there is the only force that is the gravitational force which provides centripetal force to the satellite for revolution. Therefore, the centripetal force on the satellite is equal to the gravitational force between the earth and the satellite.

Mathematically, it can be written as,

$\begin{array}{c}{F}_{g}={F}_{c}\\ \frac{G{m}_{satellite}{m}_{Earth}}{{r}^{2}}={m}_{satellite}{a}_{c}\\ {a}_{c}=\frac{G{m}_{Earth}}{{r}^{2}}\end{array}$

The relation between centripetal acceleration and speed is given by

${a}_{c}=\frac{{v}^{2}}{r}$

Now, putting the value of centripetal acceleration and calculating the speed

$\begin{array}{c}\frac{{v}^{2}}{r}=\frac{G{m}_{Earth}}{{r}^{2}}\\ v=\sqrt{\frac{G{m}_{Earth}}{r}}\end{array}$

(a)

- The mass of the Earth ${m}_{Earth}=5.97\times {10}^{24}\text{\hspace{0.33em}kg}$.
- The orbital radius of the satellite $r=6.38\times {10}^{3}\text{\hspace{0.33em}km}+900\text{\hspace{0.33em}km}=\left(6.38\times {10}^{6}+900,000\right)\text{\hspace{0.33em}m}=\text{7280000\hspace{0.33em}m}$.

So, the velocity of the satellite,

$\begin{array}{c}v=\sqrt{\frac{G{m}_{Earth}}{r}}\\ =\sqrt{\frac{6.673\times {10}^{-11}\times 5.97\times {10}^{24}}{\text{7280000}}}\\ v=7.4\times {10}^{3}\text{\hspace{0.33em}}\frac{\text{m}}{\text{s}}\end{array}$

Hence, the speed of a satellite in an orbit 900km above Earth’s surface is ${\text{7.4\xd710}}^{\text{3}}\frac{\text{m}}{\text{s}}$.

**(b)**

- The satellite's velocity w.r.t earth, ${\text{v}}_{\text{se}}{\text{=7.4\xd710}}^{\text{3}}\text{\hspace{0.33em}}\frac{\text{m}}{\text{s}}$.
- The relative velocity of the rivet w.r.t earth, ${\text{v}}_{\text{re}}{\text{=7.4\xd710}}^{\text{3}}\text{\hspace{0.33em}}\frac{\text{m}}{\text{s}}$.
- The relative velocity of the rivet w.r.t satellite, ${\text{v}}_{\text{rs}}\text{=?}$.

The satellite's velocity w.r.t. the earth and the velocity of the rivet w.r.t. the earth are perpendicular to each other, so the velocity of the rivet w.r.t. the satellite is,

$\begin{array}{c}{v}_{rs}=\sqrt{{v}_{Se}^{2}+{v}_{re}^{2}}\\ {v}_{rs}=\sqrt{{\left(7.4\times {10}^{3}\right)}^{2}+{\left(7.4\times {10}^{3}\right)}^{2}}\\ {v}_{rs}=1.05\times {10}^{4}\text{\hspace{0.33em}}\frac{\text{m}}{\text{s}}\end{array}$

Hence, the velocity of the rivet w.r.t. the satellite is ${\text{1.05\xd710}}^{\text{4}}\text{\hspace{0.33em}}\frac{\text{m}}{\text{s}}$.

**(c)**

- Given the size of the rivet, $d=3\text{\hspace{0.33em}mm}=3\times {10}^{-3}\text{\hspace{0.33em}m}$.
- The velocity of the rivet with respect to the satellite ${v}_{rs}=1.05\times {10}^{4}\text{\hspace{0.33em}}\frac{\text{m}}{\text{s}}$.

Time in which collision will last,

$\begin{array}{c}t=\frac{d}{{v}_{rs}}\\ =\frac{3\times {10}^{-3}}{1.05\times {10}^{4}}\text{\hspace{0.33em}s}\\ t=2.86\times {10}^{-7}\text{\hspace{0.33em}s}\end{array}$

Hence, the time in which collision will last is ${\text{2.86\xd710}}^{\text{-7}}\text{\hspace{0.33em}s}$.

**(d)**

- Mass of the rivet $M=0.500\text{\hspace{0.33em}g}=0.5\times {10}^{-3}\text{\hspace{0.33em}kg}$.
- The velocity of the rivet w.r.t. the satellite, ${v}_{rs}=1.05\times {10}^{4}\text{\hspace{0.33em}}\frac{\text{m}}{\text{s}}$.
- Time in which collision lasts $t=2.86\times {10}^{-7}\text{\hspace{0.33em}s}$.

The average force which rivet exerts on the satellite is,

$\begin{array}{l}F=Ma\\ F=M\frac{{v}_{rs}}{t}\\ F=\frac{0.5\times {10}^{-3}\times 1.05\times {10}^{4}}{2.86\times {10}^{-7}}\text{\hspace{0.33em}N}\\ F=1.83\times {10}^{7}\text{\hspace{0.33em}N}\end{array}$

Hence, the average force that the rivet exerts on the satellite is ${\text{1.83\xd710}}^{\text{7}}\text{\hspace{0.33em}N}$.

**(e)**

** **

Energy generated in the collision is,

$\begin{array}{l}E=\frac{1}{2}{M}_{rivet}{{v}_{rs}}^{2}\\ E=\frac{1}{2}\times 0.5\times {10}^{-3}\times {\left(1.05\times {10}^{4}\right)}^{2}\\ E=2.756\times {10}^{4}\text{\hspace{0.33em}J}\end{array}$

Hence, the energy generated in the collision is ${\text{2.756\xd710}}^{\text{4}}\text{\hspace{0.33em}J}$.

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