Does the distance to geosynchronous orbit depend on the mass of the satellite in question? Why or why not?

T = approximately 24 hs.

Explanation:

In order to keep the satellite over a fixed point on the equator, as the earth rotates, the satellite must have the same angular velocity that Earth has, which means that it must have a period equal to the time used by Earth to complete an entire rotation on itself, which is almost exactly 24 Hs.

Mathematically, this can be obtained taking into account that the force that keeps the satellite in orbit is the centripetal force, which is actually the gravitational force exerted by Earth, so we can write the following equality:

Fg = Fc ⇒ G*ms*me / (re +rsat)² = ms*ω²*(re +rsat)

By definition, ω =ΔФ / Δt

For a complete revolution, ΔФ = 2*π, and Δt = T (period of the rotation),

so we can replace ω by (2*π/T), solving then for T:

T= 86,313 sec. (24 hs are exactly 86,400 sec, so the value is actually very close to the theorical one).

Explanation:

"i have always loved contemporary art and made it a general goal to work within the field. after a few nonprofit internships at museums and galleries, i decided to try the for-profit art world and was offered a job as the ad assistant at the magazine. i much prefer the pace of the for-profit art community and have stayed with the company for three years, with a promotion to an account manager last spring."

i just did this and got a 100.

one person pulls 90n one way and the other has 100n pulling the other way.

so the net force = 90 - 100 = -10n,

or the ans can be 10n if the picture shows -90n and 100n instead.