Can the Earth orbit the Sun in just one day?

The radius of the sun rs is about 700,000 km = 700,000,000 m. The acceleration of gravity at the surface of the sun is gs = 274 m/s2.

In order for the Earth to orbit briefly above the sun’s surface in a stable orbit, there must be a balance of forces.

Centrifugal force must be equal to attraction:

Me * gs = w2 * rs * Me

with Me = Earth Mass and w = Angular Velocity.

As you can see, the earth’s mass is shortened and it remains:

gs = w2 * rs

=> w = s (gs / rs) = 6.25 * 10x-4 1/s

w = 2 * * f

with f = frequency in Hertz

f = w / 2 / . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

The orbital period U is then the inverse of the frequency:

U = 1 / f = approx. 10,000 s = approx. 2 h 47 m.

This means that the Earth could orbit the sun in about 2 hours and 47 minutes.Such an orbit would have the advantage that we no longer need to worry about greenhouse gases and global warming.

But stop.Is that really true? We have not taken into account the tidal forces of the sun. The inside of the earth is attracted to the sun a little more than the outside. If this difference is greater than the attraction of the earth, it is torn apart.

The distance at which one body is torn by another is called the Roche border 鈥?Wikipedia.

For the sake of simplicity, I read from the table and find for the Earth-Sun system the values 554,441 km, for an earth assumed as a solid and 1,066,266 km for an earth assumed to be completely liquid.

554,441 km would be within the sun, but the earth is not a solid, even if it may seem to us.It is liquid inside, with a thin crust infected with carbon units that believe they are intelligent. The Roche limit will therefore be more likely to be the higher value.

That’s why I take the Roche limit as rr = 1000,000 km to be on the safe side.

The attraction of the sun decreases with the square of the distance, which means that the attraction gr at rr

gr = gs * (rs / rr)2 = 134 m/s2

gr = w2 * rr

results.

=> w = s (gr / rr) = 3.66 * 10×4 1/s

w = 2 * * f

with f = frequency in Hertz

f = w / 2 / . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

The orbital period U is then the inverse of the frequency:

U = 1 / f = approx. 17.164 s = approx. 4 h 46m.

The Earth could therefore orbit the sun in about 4 hours 46 minutes without being torn apart by the sun.

At what distance would the Earth have to orbit the sun to take exactly one day?

gu = gs * (rs / ru)2

gu = w2 * ru

<=> gs * (rs / ru)2 = w2 * ru

<=> gs * rs2 / ru2 = w2 * ru

<=> gs * rs2 / w2 = ru3

w = 2 * * f

f = 1 / 24 h = 1.1574 * 10″-5 1/s

=> w = 7.27 * 10x-5 1/s

This results in ru to:

ru = (gs * rs2 / w2) ( 1/3) = approx. 2,939,000,000 m = 2,939,000 km

The Earth could orbit the sun at a distance of about 2.9 million km in a day.

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