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Q6.6-50PE

Expert-verifiedFound in: Page 224

Book edition
1st Edition

Author(s)
Paul Peter Urone

Pages
1272 pages

ISBN
9781938168000

**On February 14****, 2000 the NEAR spacecraft was successfully inserted into orbit around Eros, becoming the first artificial satellite of an asteroid. Construct a problem in which you determine the orbital speed for a satellite near Eros. You will need to find the mass of the asteroid and consider such things as a safe distance for the orbit. Although Eros is not spherical, calculate the acceleration due to gravity on its surface at a point an average distance from its center of mass. Your instructor may also wish to have you calculate the escape velocity from this point on Eros.**

- the centripetal acceleration is ${\text{2.08\xd710}}^{\text{-5}}\text{\hspace{0.33em}}\frac{\text{m}}{{\text{s}}^{\text{2}}}$.
- The orbital speed near Eros is $\text{1.76\hspace{0.33em}}\frac{\text{m}}{\text{s}}$.

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 *m*_{1 }and *m*_{2} are the masses of two bodies, respectively.

In angular motion, the centripetal acceleration is given,

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

As the satellite is revolving near Eros, so there is the only force that is the gravitational force which provides centripetal force to the satellite for revolution. As a result, the centripetal force of the satellite equals the gravitational force between the satellite and the entire asteroid. Mathematically, it can be written as,

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

- Mass the Eros, ${m}_{Eros}\approx 7\times {10}^{15}\text{\hspace{0.33em}kg}$.
- The orbital radius of the satellite, $r=150\text{\hspace{0.33em}km}=150,000\text{\hspace{0.33em}m}$.

So, the centripetal acceleration due to gravity on its surface,

$\begin{array}{c}{a}_{c}=\frac{G{m}_{Eros}}{{r}^{2}}\\ =\frac{6.673\times {10}^{-11}\times 7\times {10}^{15}}{{\left(150,000\right)}^{2}}\\ {a}_{c}=2.08\times {10}^{-5}\text{\hspace{0.33em}}\frac{\text{m}}{{\text{s}}^{\text{2}}}\end{array}$

Hence, the centripetal acceleration is ${\text{2.08\xd710}}^{\text{-5}}\text{\hspace{0.33em}}\frac{\text{m}}{{\text{s}}^{\text{2}}}$.

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

$\begin{array}{c}\frac{{v}^{2}}{r}=\frac{G{m}_{Eros}}{{r}^{2}}\\ v=\sqrt{\frac{G{m}_{Eros}}{r}}\\ v=\sqrt{\frac{6.673\times {10}^{-11}\times 7\times {10}^{15}}{150,000}}\\ v=1.76\text{\hspace{0.33em}}\frac{\text{m}}{\text{s}}\end{array}$

Hence, the orbital speed near Eros is $\text{1.76\hspace{0.33em}}\frac{\text{m}}{\text{s}}$.

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