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WARRIOR [948]
3 years ago
9

A large solar panel on a spacecraft in Earth orbit produces 1.0 kW of power when the panel is turned toward the sun. What power

would the solar cell produce if the spacecraft were in orbit around Saturn, 9.5 times as far from the sun?
Physics
1 answer:
Mandarinka [93]3 years ago
3 0

Answer:

e*P_s = 11 W

Explanation:

Given:

- e*P = 1.0 KW

- r_s = 9.5*r_e

- e is the efficiency of the panels

Find:

What power would the solar cell produce if the spacecraft were in orbit around Saturn

Solution:

- We use the relation between the intensity I and distance of light:

                                  I_1 / I_2 = ( r_2 / r_1 ) ^2

- The intensity of sun light at Saturn's orbit can be expressed as:

                                  I_s = I_e * ( r_e / r_s ) ^2

                                  I_s = ( 1.0 KW / e*a) * ( 1 / 9.5 )^2

                                  I_s = 11 W / e*a

- We know that P = I*a, hence we have:

                                  P_s = I_s*a

                                  P_s = 11 W / e

Hence,                       e*P_s = 11 W

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Let <em>B</em> = <em>a</em> <em>i</em> + <em>b</em> <em>j</em> + <em>c</em> <em>k</em>. Then the cross product of <em>A</em> = <em>i</em> + 2<em>j</em> - <em>k</em> with <em>B</em> is

<em>A</em> × <em>B</em> = ( <em>i</em> + 2<em>j</em> - <em>k </em>) × ( <em>a</em> <em>i</em> + <em>b</em> <em>j</em> + <em>c</em> <em>k</em> )

<em>A</em> × <em>B</em>  = <em>a</em> ( <em>i</em> × <em>i</em> ) + 2<em>a</em> ( <em>j</em> × <em>i</em> ) - <em>a</em> ( <em>k</em> × <em>i </em>)

… … … + <em>b</em> ( <em>i</em> × <em>j</em> ) + 2<em>b</em> ( <em>j </em>× <em>j</em> ) - <em>b</em> ( <em>k</em> × <em>j</em> )

… … … + <em>c</em> ( <em>i</em> × <em>k</em> ) + 2<em>c</em> ( <em>j</em> × <em>k</em> ) - <em>c</em> ( <em>k</em> × <em>k</em> )

<em>A</em> × <em>B</em> = 0 - 2<em>a</em> <em>k </em>- <em>a</em> <em>j</em>

… … … + <em>b</em> <em>k</em> + 0 + <em>b</em> <em>i</em>

… … … - <em>c</em> <em>j</em> + 2<em>c</em> <em>i</em> - 0

<em>A</em> × <em>B</em> = (<em>b</em> + 2<em>c</em>) <em>i</em> + (-<em>a</em> - <em>c</em>) <em>j</em> + (<em>b</em> - 2<em>a</em>) <em>k</em>

So we have

3 <em>i</em> - <em>j</em> + 5 <em>k </em>= (<em>b</em> + 2<em>c</em>) <em>i</em> + (<em>c</em> - <em>a</em>) <em>j</em> + (<em>b</em> - 2<em>a</em>) <em>k</em>

which gives us the system of equations,

{ <em>b</em> + 2<em>c</em> = 3

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Solve for <em>a</em>, <em>b</em>, and <em>c</em>.

• Eliminate <em>c</em> from the first two equations:

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Answer:

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Have a nice day!          

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Answer:

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