Question

Prine The New Horizons space probe transmits to the earth at a frequency of 8.3 GHz, At some point the intensity of the wave was I = 2.85e-25 W / m2

a) What will be the intensity of the wave three times the spacecraft?

b) What is the wavelength of the transmitted signal?

If the power of the sun is 3.88e + 26 w.  

c) At what distance from the sun, the radiation intensity will be 0.263 W / m²?

d) What will be the absorption radiation pressure exerted by the sun on the probe at that distance?

e) If we consider the above distance as the New Horizons distance to the earth. And that the power of the antenna is 17 W. What will be the intensity of the wave transmitted by New horizons that reaches the earth?

f) How many times farther is the New Horizons probe from the beginning of the problem?

Answer 1

a) Intensity is inversly proportional to square of the distance.

so, The intensity of the wave three times the spacecraft = 3*2.85*10^-25

= 8.55*10^-25 W/m^2

Note : If we need intensity at three times of initial distance, I = 2.85*10^-25/3^2

= 3.17*10^-26 W/m^2

b) wavelength, lamda = c/f

= 3*10^8/(8.3*10^9)

= 0.0361 m

c) let r is the distance where intensity is 0.263 W/m^2

use, I = P/(4*pi*r^2)

r^2 = P/(4*pi*I)

r = sqrt(P/(4*pi*I))

= sqrt(3.88*10^26/(4*pi*0.263))

= 1.08*10^13 m

d) Obsorotion radiation pressure = I/c

= 0.263/(3*10^8)

= 8.77*10^-10 Pa

e) Intensity, I = P/(4*pi*r^2)

= 17/(4*pi*(1.08*10^13)^2)

= 1.16*10^-26 W/m^2

f) let I1 = 2.85*10^-25 W/m^2
let r1 is the initial distance
I2 = 1.16*10^-26 W/m^2

new horizon distance, r2 = ?

use, I2/I1 = (r1/r2)^2

sqrt(I2/I1) = r1/r2

r2 = r1*sqrt(I1/I2)

= r1*sqrt(2.85*10^-25/(1.16*10^-26))

= r1*4.96

so, 4.96 times farther is the New Horizons probe from the beginning of the problem

Note : please comment for any further clarification.