# Satellite Motion centripetal force

Once I read in a book that the centripetal force that keeps the satellite in its circular orbit around the Earth is approximately equal to its weight. My question is why the word "approximately" was used in this sentence? Do the forces really differe or it is just a dictation error?

If we call "weight" the gravitational force that earth makes over a body (by example, the weight of a person in the surface, equal to $$mg$$), this definition can also be used to describe the gravitational force over the satellite. This force is the one that allows it to continue its circular orbit, the centripetal force. From this point of view, the centripetal force is not "approximate" equal to the weight, is exactly equal to the weight.

A different issue is that the gravitational force over the satellite (weight) when in its orbit is different of the one when the satellite was in the surface. Depending on the orbit height (see here) the difference between these two weights could range from 5% to up to 95%. By example, at ISS (International Space Station) gravitational force is around 90% of the surface one.

• +1 for a good answer. I believe the right notion of weight is that it always equals the gravitational force, as you mention first. Even when far away from the surface. When on the Moon, we would also say that the weight is smaller. It is equal to the gravitational force where ever you are. – Steeven Jan 27 '19 at 20:25
• Which satellites have an orbit with an altitude of 30 km? – noah Jan 27 '19 at 20:58
• @pasabaporaqui From your link: "Low Earth orbit (LEO): geocentric orbits with altitudes from 160 to 2,000 km" – noah Jan 27 '19 at 21:04
• Note that usual definition of weight is the sum of all inertial forces including gravity in the current reference frame. Of course, the centripetal force will be exactly equal to weight in whichever reference frame we chose (notably either rotating with Earth, or non-rotating) as there are no other forces in the relevant direction. – Jan Hudec Jan 27 '19 at 21:55

Probably the word "approximately" is used because the term "weight" is ambiguous when talking about a satellite in orbit. If you weighed the satellite at the top of a 100-mile tall tower, it would weigh a bit less than it weighs on the ground, because the strength of gravity falls as altitude increases. The satellite's mass does not change, but its weight changes.

$$mv^2/r$$ = $$GmM/r^2$$ as you would know for the satellite in orbit around the earth, with $$m$$ being the mass of satellite, $$M$$ as that of earth and $$r$$ is the distance from the center of the earth to the satellite in space.

$$r=R+h$$ where $$R$$ is the radius of the earth. So, if you were about in your daily life measuring your weight ($$mg$$), the $$g=9.8 m/s^2$$ is the value on the surface. In space, $$h$$, which is the height above the surface factors in.

$$g_n =g (1+\frac{h}{R})^{-2}$$

with $$g_n$$ being the new gravitational acceleration term. So the resulting weight would be $$mg_n\approx mg$$ as $$g_n \approx g$$ (this mostly depends on where you keep your satellite - this "approximation" fails if that height $$h$$ is comparable to $$R$$ - that is of the same order of magnitude).

Satellites that realistically fit this approximation (for your information) like space telescopes which orbit at a height of about 300 miles above the surface.

• Which is the relation between first formula and the question ? $v$ seems unrelated to it and to the remainder of your answer. – pasaba por aqui Jan 27 '19 at 20:04
• It was just to highlight the centripetal nature of the gravitational force. – KV18 Jan 27 '19 at 20:39