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
Answer:
Our solar system has total eight planets out of which four are inner planets and four are outer planets. The four outer planets are Jupiter, Saturn, Uranus and Neptune. The common characteristics of outer planets is that they are gaseous planets. They are larger on size than the inner rocky planets and are faraway from Sun. They have larger period of revolution around the Sun.
Uranus is a gaseous planet and lies far from Sun and hence has large period of revolution. It takes 84 Earth years to revolve around Sun. This data indicates that Uranus resides in the outer region of the Solar System.
Answer:
a. 8.33 x 10 ⁻⁶ Pa
b. 8.19 x 10 ⁻¹¹ atm
c. 1.65 x 10 ⁻¹⁰ atm
d. 2.778 x 10 ⁻¹⁴ kg / m²
Explanation:
Given:
a.
I = 2500 W / m² , us = 3.0 x 10 ⁸ m /s
P rad = I / us
P rad = 2500 W / m² / 3.0 x 10 ⁸ m/s
P rad = 8.33 x 10 ⁻⁶ Pa
b.
P rad = 8.33 x 10 ⁻⁶ Pa *[ 9.8 x 10 ⁻⁶ atm / 1 Pa ]
P rad = 8.19 x 10 ⁻¹¹ atm
c.
P rad = 2 * I / us = ( 2 * 2500 w / m²) / [ 3.0 x 10 ⁸ m /s ]
P rad = 1.67 x 10 ⁻⁵ Pa
P₁ = 1.013 x 10 ⁵ Pa /atm
P rad = 1.67 x 10 ⁻⁵ Pa / 1.013 x 10 ⁵ Pa /atm = 1.65 x 10 ⁻¹⁰ atm
d.
P rad = I / us
ΔP / Δt = I / C² = [ 2500 w / m² ] / ( 3.0 x 10 ⁸ m/s)²
ΔP / Δt = 2.778 x 10 ⁻¹⁴ kg / m²
