Book edition 1st Edition

Author(s) Paul Peter Urone

Pages 1272 pages

ISBN 9781938168000

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Short Answer

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 ${2.08×10}^{-5} \frac{m}{s^{2}}$ .
The orbital speed near Eros is $1.76 \frac{m}{s}$ .

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Step by Step Solution

TABLE OF CONTENTS :

TABLE OF CONTENTS

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 m₁and m₂ are the masses of two bodies, respectively.

In angular motion, the centripetal acceleration is given,

$\begin{matrix} F_{c} = m a_{c} \\ F_{c} = m ω^{2} r \end{matrix}$

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,

$\begin{matrix} F_{g} = F_{c} \\ \frac{G m_{s a t e l l i t e} m_{E r o s}}{r^{2}} = m_{s a t e l l i t e} a_{c} \\ a_{c} = \frac{G m_{E r o s}}{r^{2}} \end{matrix}$

Step 2: Given data

Mass the Eros, $m_{E r o s} \approx 7 \times 10^{15} kg$ .
The orbital radius of the satellite, $r = 150 km = 150, 000 m$ .

Step 3: Centripetal acceleration of the satellite

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

$\begin{matrix} a_{c} = \frac{G m_{E r o s}}{r^{2}} \\ = \frac{6.673 \times 10^{- 11} \times 7 \times 10^{15}}{{(150, 000)}^{2}} \\ a_{c} = 2.08 \times 10^{- 5} \frac{m}{s^{2}} \end{matrix}$

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

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

$\begin{matrix} \frac{v^{2}}{r} = \frac{G m_{E r o s}}{r^{2}} \\ v = \sqrt{\frac{G m_{E r o s}}{r}} \\ v = \sqrt{\frac{6.673 \times 10^{- 11} \times 7 \times 10^{15}}{150, 000}} \\ v = 1.76 \frac{m}{s} \end{matrix}$

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

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College Physics (Urone)

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Short Answer

Step by Step Solution

Step 1: Centripetal acceleration of the satellite

Step 2: Given data

Step 3: Centripetal acceleration of the satellite

Step 4: Determining the orbital speed near Eros

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College Physics (Urone)

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Short Answer

Step by Step Solution

Step 1: Centripetal acceleration of the satellite

Step 2: Given data

Step 3: Centripetal acceleration of the satellite

Step 4: Determining the orbital speed near Eros

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