Let 10 re be the distance of the satellite a from the center of the earth, where re is the radius of the earth. what is the gravitational acceleration due to the earth at satellite a?

a = 0.098 m/s²

Explanation:

The satellite, at any distance from the center of the Earth, is subject to the attractive force due to the Earth, according to the Newton´s Universal Law of gravitation, as follows:

Fg = G*ms*mE / (rse)²

According to Newton´s 2nd Law, neglecting any other force acted upon the satellite, we can write the following equation:

Fg = ms*a = G*ms*mE / (rse)²

⇒ a = G*mE / (rse)² (1)

As the distance between the satellite and the center of the Earth is 10 times the radius of the Earth, replacing this value in (1), we have:

a = G*mE / (10*RE)² = G*mE/(RE)² * (1/100)

but G*mE/(RE)², is just g, the acceleration due to gravity on the surface of the earth, so the gravitational acceleration due to Earth at satellite A, is as follows:

a = g*(1/100) = 0.01*g = 0.098 m/s²

most voltage drop calculators want the distance between the source and destination. then internally, they double it so both the source and return wires are calculated. but you need to know which distance the calculator uses so you're not half of what you should be or double what you should be.

most circuit breakers will not take a #6 directly, so the electrician will have to splice on a smaller pigtail. you also have to upsize the grounding wire when you upsized the ungrounded (hot) wire (see note in 250.122). this will preclude using most cable assemblies, as a cable with #6 copper will most likely only have a #10 grounding conductor. you need a #6 grounding conductor if the circuit is 20a. a #8 cable is no better, as it will have a #10 grounding conductor too. but most 20a breakers will take a #8 directly.

in your related post, you didn't say what type of load this is. if it is a light or heater, voltage drop won't be a big issue. if this is a motor, the current needs are huge at startup and the voltage drop can pose a problem. if this is a motor, knowing if the motor is under much load during startup will answer the voltage drop effect (a saw is easier to start than a compressor).