What fraction is this of the satellite's weight at the surface of the earth? take the free-fall acceleration at the surface of t

Question

MariettaO

29 days ago

7

What fraction is this of the satellite's weight at the surface of the earth? take the free-fall acceleration at the surface of t

he earth to be g = 9.80 m/s2?

Physics

1 answer:

kicyunya [3.2K] · Answer 1 · 2026-03-31T16:14:12+03:00

I have included the additional details regarding the question.
It is important to remember that weight represents the strength of the gravitational pull on an object.
The weight at the surface can be expressed as:
$F_g=mg$
Here, we utilize the standard g value. However, g varies with elevation. The reason this variation is rarely observed is due to the fact that the height must be considerable relative to Earth's radius for it to make a notable impact.
According to Newton's law of gravitation, we have:
$F_g=G\frac{m_em}{r^2}$
In this scenario, r refers to the distance between the centers of mass of the interacting objects. While standing on the Earth's surface, r is equivalent to 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 denote as g.
When at a certain height, the expression is:
$F_g=G\frac{m_em}{(r_e+h)^2}\\ ma=G\frac{m_em}{(r_e+h)^2}\\ a=G\frac{m_e}{(r_e+h)^2}\\$
Let’s call this new acceleration g'. Dividing it by g gives us:
$\frac{g'}{g}=\frac{r_e^2}{(r_e+h)^2}\\ g'=g\cdot\frac{r_e^2}{(r_e+h)^2}$
Thus, the weight can be expressed as:
$F_g'=mg'=mg\cdot\frac{r_e^2}{(r_e+h)^2}$
By substituting the values, we find:
$F_g'=2.02N$
For Earth, we obtain:
$F_g=22050N$
Consequently, the resulting ratio is:
$\frac{F_g'}{Fg}=\frac{2.02}{22050}=0.000091=0.0091\%$