So, is it correct to measure centrafugal force due to a planet's rapid orbit?

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The discussion centers on the measurement of centrifugal force due to a planet's rapid orbit, specifically focusing on Corot-7b, which has a diameter 1.7 times that of Earth and orbits its star in just 20 hours and 29 minutes. The centrifugal force experienced on Corot-7b reaches an extreme 14.8g, while the planet's gravity is slightly higher at 16g. This results in significant gravitational discrepancies across the planet's surface, with the dayside experiencing 30.8g and the backside 1.2g. The conversation also highlights the formula used for calculating gravity from centrifugal force: g=R*(pi*rpm/30)^2.

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Hello
i was concerned about how fast should Earth spin so that on equator gravity would be 0 due to centrafugal force i got 84 minutes - full spin

since it is more like 24 hours than on equator currently centrafugal force is 0,039g
but there is also one smaller centrafugal force coming from orbiting the sun and it is 0.005g
non feelable. barely measurable...

the problem is that the closer you get to host star the biger the centrafugal force from rotating the star become. for mercury it is 0,04g

but there is also a smallest known planet corot-7b with 1.7 Earth diameters slightly denser and 23 times closer to its star than mercury to sun. which makes it have its full year ONLY in 20hours and 29 minutes
it turns out that spining so fast makes a centrafugal effect of 14,8g which is abolutelly enormous well the planets gravity is still slightly bigger 16g
but this fact is bringing in some weird consequences

1. the planet is most likely face locked.
2. the gravity at the backside is 1.2 g
3. the gravity at the front side is 30.8g WOW
4. the funniest thing is what the gravity at the sides are
it is actually somewhere inbetween but it is not pointing down but at an angel away from star.
so it would feel like climbing down a hill. now if you happen to have a bicycle and you are at the dayside . you can always take a ride to night side just releasing the brakes.

is it correct measuring centrafugal force due to planets rapid orbit.

i used formula for calculating gravity from centrafugal force like this
g=R*(pi*rpm/30)^2 where R-radius in meters, rpm -ratePerMinute
 
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If a planet is in orbit around a sun, there is no net "outward force". Remember there is the gravity of the Sun to account for. It is like the object sitting at the equator with the Earth rotating in 84 min. The tendency to fly off is canceled by the gravity pulling in.

That is not to say that there would not be a difference between the Sun facing side and its opposite, but this would be due to their different distances from the Sun. For Corot-7b this works out to a difference of less than 2/100 g.
 

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