Moving the Earth 7.5 meters in one year against a force of 3.5e18 Kg requires 8e11 joules which is 8e11 watts.
Wouldn't that be 8e11 joules / 1 year to get watts? That would make the power needed = 8e11 joules / (1 year x 365 days/year x 24 hours/day x 3600 seconds/hour) = 25,000 joules/second = 25,000 watts. Watts is power, which is energy per unit time.
That doesn't seem like a lot of power, since I have a 25,000 watt generator.
But, wait, there are problems with the 8x10^11 joules number, too.
Going back a bit further, Kg is mass, not force, so the force result should be in Kg-m/sec^2 which is "newtons" of force. Using Bill's numerical values for G, the mass of the Sun and Earth, and 93 million miles for the radius of Earths orbit, I get 3.62 x 10^22 Kg-m/sec^2. So, I think Bills force estimate should be 3.62 x10^22 Newtons of force. And then moving the Earth 7.5 meters against that force would take 7.5 x 3.62x10^22 = 2.72x10^23 newton-meters of energy. A newton-meter is a Joule of energy.
So, my number is much higher than Bill's for the energy needed.
So, 2.72x10^23 Joules of energy over a year is still a continuous power level of 8.6x10^15 watts, which would be necessary to be applied with 100% efficiency to move the Earth that 7.5 meters farther away from the sun over a years time.
For comparison, global electric power generation in 2021 was 28.2 terawatt-hours = 2.82x10^13 watt-hours = 3.22x10^9 watt-years. So, that is about 3.22x10^9 / 8.6x10^15 = 0.000037 of the power needed to move the Earth 7.5 meters farther from the sun over a year.
And, that doesn't begin to address the question of
how to apply that power to produce motion of the Earth. We would have to eject mass from the Earth at high velocity, which is far from 100% efficient if we are talking about launching weights to escape velocity.