(a) [tex]1.15\cdot 10^9 N[/tex]
For an electromagnetic wave incident on a surface, the radiation pressure is given by (assuming all the radiation is absorbed)
[tex]p=\frac{I}{c}[/tex]
where
I is the intensity
c is the speed of light
In this problem, [tex]I=1360 W/m^2[/tex]; substituting this value, we find the radiation pressure:
[tex]p=\frac{1360 W/m^2}{3\cdot 10^8 m/s}=4.5\cdot 10^{-6} Pa[/tex]
the force exerted on the Earth depends on the surface considered. Assuming that the sunlight hits half of the Earth's surface (the half illuminated by the Sun), we have to consider the area of a hemisphere, which is
[tex]A=2pi R^2[/tex]
where
[tex]R=6.37\cdot 10^6 m[/tex]
is the Earth's radius. Substituting,
[tex]A=2\pi (6.37\cdot 10^6 m)^2=2.55\cdot 10^{14}m^2[/tex]
And so the force exerted by the sunlight is
[tex]F=pA=(4.5\cdot 10^{-6} Pa)(2.55\cdot 10^{14} m^2)=1.15\cdot 10^9 N[/tex]
(b) [tex]3.2\cdot 10^{-14}[/tex]
The gravitational force exerted by the Sun on the Earth is
[tex]F=G\frac{Mm}{r^2}[/tex]
where
G is the gravitational constant
[tex]M=1.99\cdot 10^{30}kg[/tex] is the Sun's mass
[tex]m=5.98\cdot 10^{24}kg[/tex] is the Earth's mass
[tex]r=1.49\cdot 10^{11} m[/tex] is the distance between the Sun and the Earth
Substituting,
[tex]F=(6.67\cdot 10^{-11} )\frac{(1.99\cdot 10^{30}kg)(5.98\cdot 10^{24} kg)}{(1.49\cdot 10^{11} m)^2}=3.58\cdot 10^{22}N[/tex]
And so, the radiation pressure force on Earth as a fraction of the sun's gravitational force on Earth is
[tex]\frac{1.15\cdot 10^9 N}{3.58\cdot 10^{22}N}=3.2\cdot 10^{-14}[/tex]
