Question

What fraction is this of the satellite's weight at the surface of the earth? take the free-fall acceleration at the surface of the earth to be g = 9.80 m/s2?

Answer 1

I attached the missing information about the question.
You must keep in mind that weight is simply the magnitude of the gravitational force acting on the object.
We know that weight on the surface is simply:
$F_g=mg$
Where we use that standard value of g. Hoewer g changes with height. The reason why people don't see this that often is because that height has to be relatively big compared to the Earth's radius in order to make a significant difference.
Newton's law of gravity states:
$F_g=G\frac{m_em}{r^2}$
Where r is the distance between centers of mass of two objects that are interacting. When you are standing on the ground r is simply Earth's radius:
$F_g'=G\frac{m_em}{r_e^2}\\ ma=G\frac{m_em}{r_e^2}\\ a=G\frac{m_e}{r_e^2}$
This acceleration is what we call g.
Now, when you are at some height you have:
$F_g=G\frac{m_em}{(r_e+h)^2}\\ ma=G\frac{m_em}{(r_e+h)^2}\\ a=G\frac{m_e}{(r_e+h)^2}$
Let's call this acceleration g'. We divide it by g, and we get:
$\frac{g'}{g}=\frac{r_e^2}{(r_e+h)^2}\\ g'=g\cdot\frac{r_e^2}{(r_e+h)^2}$
Now the weight is simply:
$F_g'=mg'=mg\cdot\frac{r_e^2}{(r_e+h)^2}$
When we plug in the numbers we get:
$F_g'=2.02N$
On Earth we get:
$F_g=22050N$
So the ratio is:
$\frac{F_g'}{Fg}=\frac{2.02}{22050}=0.000091=0.0091\%$