Answer:
a = 5.037 10⁻² m / s²
Explanation:
For this exercise we will use the concept of radiation pressure for a reflective surface
P = S / c
Where c is the speed of light and S is the pointing vector which is equal to the intensity of the radiation
The definition of Pressure is
P = F / A
F = P A
Newton's second law gives the relationship between force and acceleration
F = m a
We substitute
P A = m a
a = I A / c m
Let's look for the radiation intensity, for this we use that the intensity in the Earth's orbit is 1360 W / m².
We must look for the intensity in the orbit would want 0.5 AU, for this we use the definition of instance
I = P / A
Where P is the power of the sun's emission that is constant
P = I A
I₁ A₁ = I₂ A₂
Where index 1 is for Earth and 2 for the orbit at 05 AU
he area of a sphere is
A = 4 π R²
I₂ = I₁ A₁ / A₂
I₂ = I₁ R₁² / R₂²
I₂ = 1360 (150 10⁶/75 10⁶)²
I₂ = 5440 W
This is the radiation at the point of interest
Let's calculate
a = 5440/3 10⁸ 10⁶/360
a = 5.037 10⁻² m / s²
