Good Bye Gravity!

Suppose you are standing on the surface of the earth and suddenly, the attraction due to gravity exerted by the earth vanishes. What would your position be (for example, you would fly up, fly tangentially or remain standing, etc) just at the instant after the attraction has vanished? Also, what would be your position if the revolution of the earth is neglected?

Note by Vaishnavi Gupta
6 years, 11 months ago

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as there is no resultant force after the gravity vanishes you would remain no need to worry in saying bye to gravity

somesh ROUT - 6 years, 11 months ago

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Well, that does seem quite right but I was thinking that a person standing on the surface of the earth would tend to fly tangentially, in the case where the revolution of the earth is taken into consideration. Its something similar to the situation when you tie a stone at the end of a string and whirl it in a circle but if the string is suddenly cut, the stone flies away tangentially due to inertia of direction. What do you think about this possibility?

Vaishnavi G - 6 years, 11 months ago

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not revolution, but rotation. So, you will orbit the sun instead and you will meet earth every year.

Fahad Shihab - 6 years, 10 months ago

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I agree. Although this factor would be insignificant, as some other force (wind, jumping, breathing hard) would disturb you by the time you would be able to notice that effect.

Bob Krueger - 6 years, 11 months ago

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@Bob Krueger well yeah...but what if we consider some ideal situation, ignoring these effects( breathing hard, wind, etc)....what EXACTLY would happen to that man, then? [Still haven't got a satisfactory answer from anyone :( ]

Vaishnavi Gupta - 6 years, 11 months ago

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@Vaishnavi Gupta If you neglect everything like buildings,trees,air drag,etc. and assume the Earth to be a completely barren sphere, then -

Just before the gravitational force became 0, you were at rest w.r.t. the Earth. Thus w.r.t. a person at rest in space, you would be in a complicated motion-

circular due to rotation of the Earth and another bigger circular motion due to revolution of the Earth.

Thus when it becomes 0, you can't be at rest with Earth as the gravity, Normal and Friction all are 0.You would then possess the velocity which you had just before the gravity became 0. Now comes the fun part. If you were on the side facing the sun(i.e. if it was day-time when it happened) and your velocity's direction was in the path of the orbit of the Earth around the Sun(assuming the poor Sun still has its gravity), then the Earth will(or should) cancel out your path as it will move inwards(as it is following it's orbit) and so a Normal reaction force will act between you and the Earth which should propel you towards the sky(not straight upwards, but rather like the path of a reflected ray of light) and potentially towards the Sun! If it is night, then you should fly off tangentially.

This is all, of course, neglecting the gravitational pull of the Sun, the Moon and the other planets will have on you. If we take that into consideration, then it becomes much more complex (and intellectually enjoyable).

And lastly, I am not saying I am correct. Feel free to imagine better than me!

Siddharth Brahmbhatt - 6 years, 11 months ago

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@Siddharth Brahmbhatt the direction is dependant only on the position when it happened. the equations below proves it.

Fahad Shihab - 6 years, 9 months ago

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@Siddharth Brahmbhatt But wait, if you want to calculate the trajectory, use this. Assuming Rotational speed=402m/s As the rotation is 23.5degrees tilted, we should have a Δvhorizontal=cos23.5×402\Delta v_{horizontal}=\cos 23.5 \times 402

Δvvertical=sin23.5×402\Delta v_{vertical}=\sin 23.5 \times 402


a=rmin(1+e)a=r_{min}(1 + e)

vorbit=(1+e)GMsunrperihelionv_{orbit}=\sqrt{\frac{(1+e) GM_{sun}}{r_{perihelion}}}

Ang.Distance from perihelion is θ\theta



Also we want to find the speed of earth at that point. So, we can derive the equation from angular momentum' conservation.



Δωinclination=ωinclination+(cosθ2(cosθ2vverticalearth+Δvvertical))\Delta \omega_{inclination}= \omega_{inclination}+(\cos\theta_{2}(\cos\theta_{2}v_{vertical-earth}+\Delta v_{vertical}))

θ2\theta_{2} is ang.Distance from the ascent node.

Δvorbit=vcurrent+Δvhorizontal\Delta v_{orbit}= v_{current}+\Delta v_{horizontal}

now, θ3\theta_{3} is the angle between periapsis of object. when it is below 90, the apoapsis will raise more than periapsis. But, if it is above 90, it will raise the periapsis but θ3180\theta_{3}\leq180

So, Δperiapsis=cosθ3(Δvhorizontal)\Delta periapsis=\cos\theta_{3}(\Delta v_{horizontal})

Δapoapsis=sinθ3(Δvhorizontal)\Delta apoapsis=\sin\theta_{3}(\Delta v_{horizontal})

So, with that we can find that we are actually oscillating about the orbit of earth(but slightly higher one)

How's it? Hard one? Guys! these are only the beginning of Orbital Physics!

Fahad Shihab - 6 years, 10 months ago

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No need to worry?! I think gravity is actually rather convenient.

Tim Vermeulen - 6 years, 11 months ago

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