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Q6.6-50PE

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### College Physics (Urone)

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 ${\text{2.08×10}}^{\text{-5}}\text{ }\frac{\text{m}}{{\text{s}}^{\text{2}}}$.
• The orbital speed near Eros is $\text{1.76 }\frac{\text{m}}{\text{s}}$.
See the step by step solution

## Step 1: Centripetal acceleration of the satellite

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 m1 and m2 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,

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

## Step 2: Given data

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

## Step 3: Centripetal acceleration of the satellite

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

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

Hence, the centripetal acceleration is ${\text{2.08×10}}^{\text{-5}}\text{ }\frac{\text{m}}{{\text{s}}^{\text{2}}}$.

## Step 4: Determining the orbital speed near Eros

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

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

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