Question

The average energy flux in the sunlight on the Earth is ⟨S⟩=1.4×103 W/m2. You might need to use some of the following constants:
Distance from earth to the sun=AU=150×109m
REarth=6.4×106m
RSun=7.0×108m
G=6.67×10−11m3/(kgs2)
MEarth=5.97219×1024kg
MSun=1.9891×1030kg

(a) What force (in N) does the pressure of light exert on the Earth? Assume that all the light striking the Earth is absorbed.



(b) What is the gravitational force (in N) that the Sun exerts on the Earth? (Think about how that compares to the force due to the pressure of light. Does your answer make sense?)

Answer 1

Check units: W/m^2 = Js^(-1) m^(-2)
J= N m =>W/m^2= Js^(-1)m^(-2)=Nm^(-1)s^(-1)

so W/m^2/(m/s) = Nm^(-1)s^(-1)/ms^(-1) = Nm^(-2) = pressure

so we have verified that the units of intensity/speed of light = units of radiation pressure

therefore, the force of radiation acting on the earth is:

force = radiation pressure x area = (intensity/c)xpi R^2
force = 1400W/m^2 x pi x( 6.37x10^6m)^2/3x10^8m/s
force = you do it.
I am presuming that the sun's gravitational attraction means the magnitude of the solar gravitational force on earth: If that's the case, the answer is approx 10^22 N:

F=GMm/r^2

G=6.67x10^(-11)=6.67e-11
M=mass sun = 2x10^30kg=2e30
m=mass earth = 6x10^24kg
r=earth sun distance = 1.5x10^11m

F=(6.6e-11)(2e30)(6e24)/(1.5e11)^2 = 3.56e22N

Answer 2

Thanks bobpursley!
The first problem does not give right answer, could you tell, if this is the right calculation as 10^22 and the formula was not right.

1400/6x10^24*pi*(6.37*10^6)^2/3*10^8

Is this formula above right?

Answer 3

ist part ans is 5.823e8

Answer 4

Thanks hmmmmm for help!

Answer 5

Pls other questions, 1, 2, 4?

Answer 6

B=(meu not*N*I)/((2*pi*R/km) +d)

Answer 7

this was for question 4

Answer 8

what is meu not?

Answer 9

the correct ans is:
B=(mu_0*N*I)/((2*pi*R/km) +d)

Answer 10

¿Problems 1, 2 & 5?

Answer 11

how to count 3b (Bo) and 3c (Poynting vector)?