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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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# Gravity vs centrafugal force

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