We analyse the force and power required to keep the earth in orbit from two points of view. First, the tension on a steel cable give a sense of scale of the huge forces involved. Then we look at a Pushing Gravity force which has a long and interesting history.
vt = circumference/period = 2*pi*au/year = 2*pi*149.598E9_m/31.56E6_s = 29,785.9_m/s, the tangent velocity of the earth in orbit. 149.598E9_m = au, the average distance between the centers of the earth and sun. The year is expressed in seconds.
To deflect the mass of the earth, Mer = 5.9722E24_kg, from her inertial, straight line path, requires a centripetal force of,
mass*velocity2/radius = Mer*vt2/au = 5.9722E24_kg*(29,785.9_m/s)2/149.598E9_m = 3.5418E22_kg*m/s2.
Force resisted by steel:
Ultimate strength is quoted force per unit of cross section area (N/mē). The SI unit of stress is the pascal, where 1_Pa = 1_N/m2 = 1_kg/(m*s2). The ultimate tensile strength of AISI 1018 Steel is 440 MN/m2 = 440E6_kg/(m*s2) = 63,816_psi. The earth could be held back by, centripetal force/tensile strength = area, 3.5418E22_kg*m/s2 / (440E6_kg/(m*s2)) = 8.0497E13_m2 of steel cable. The pi*radius2 area of the circle of the earth is pi*(6.378E6_m)2 = 1.2747E14_m2. The cable would need to be 63 percent of this area of the earth. The sun and earth are seen connected together by a cable like a giant dog bone or dumb bell. The cable could be replaced by a constant flux of particles or waves continuously pushing against the earth to keep it in orbit. This would be Pushing Gravity. The cable could also be replaced by opposite charges in the sun and earth. This would be Newtonian, electrostatic, action-at-a-distance gravity or electrostatic gravity.
To keep her in orbit also requires a centripetal radial velocity of,
2*Au/Year = 2*149.598E9_m/31.56E6_s = 9,481.45_m/s
Power to keep earth in orbit:
The power required to keep the earth in an orbit, to be applied each second, in the absense of gravity, would be,
force * velocity = power
centripetal force * radial velocity = Watts
Mer*vt2/Au * 2*Au/Year = 3.357E26_kg*m2/s3 or 3.357E26_W
To find the minimum mass required to generate this power, we divide by c2
3.357E26_kg*m2/s3/c2 = 3.736E9_kg/s, required each second. This is the power of a four megaton nuke per second. (2000E6_lbs = 906E6_kg. 906E6_kg*c2 = 8.15E25_W). How much would this heat-up the earth? 3.357E26_W/5.9722E24_kg = 56.21_W/kg. This is a mass energy of 1.179E17_kg/year. This is an earth mass of energy every 50.65 million years. This says the earth has used this much mass since the time of the dinosaurs just to stay in orbit.
The force required to keep the earth in orbit is provided by the impulse of unseen waves or particles with mass or energy which come in from every direction and which are stopped by everything which has gravity. Space is supposed filled with these unseen particles. Objects shadow each other from these particles and are pushed together by the impulse of the particles. The impact of these particles must generate a lot of heat.
Solar output power:
Earth's insolation = 1,366_W/m2, The solar output received at the earths orbit.
4/3*pi*au2 = 9.3743E22_m2, The area of a sphere with a radius of the earths orbit.
insolation * area = 1,366_W/m2 * 4/3*pi*au2 = 1.2805E26_W, The total solar power output, is the power incident on the area of this sphere.
Pushing gravity power required:
The power to keep the earth in orbit, by means of pushing gravity 3.357E26_W, is more than twice the total solar output of 1.2805E26_W. This is as unrealistic, one might say silly, as the steel cable.