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asd
#91

vickachu
Reply +7 123456789123345869
(07/05/2013) [] how much energy would there be needed to stop something as big as our planet from spinning?
#111 to #91

mralice
Reply +17 123456789123345869
(07/05/2013) [] Assuming you mean spinning, and not orbiting...
Newtons second law of motion is Force = (Mass)(Acceleration). F=MA.
Earths mass is roughly 5.9736 × 10^24 kg (according to Google).
The Earths acceleration is 0.03 m/s2 (soruce at tinyurl.com/fjscience).
So... 1.79208 x 10^23 Newtons of force would be needed at the exact opposite vector the Earth is spinning in.

1 N is 0.1225 lbs so you would need a force of 21,952,980,000,000,000,000,000 pounds.
Newtons second law of motion is Force = (Mass)(Acceleration). F=MA.
Earths mass is roughly 5.9736 × 10^24 kg (according to Google).
The Earths acceleration is 0.03 m/s2 (soruce at tinyurl.com/fjscience).
So... 1.79208 x 10^23 Newtons of force would be needed at the exact opposite vector the Earth is spinning in.

1 N is 0.1225 lbs so you would need a force of 21,952,980,000,000,000,000,000 pounds.
#243 to #111

anon
Reply 0 123456789123345869
(07/05/2013) [] Almost.
This problem requires rotational mechanics, i.e. mass is replaced with moment of inertia, force by moment and momentum but angular momentum. Besides, stopping linear movement by applying a force also requires time.
But to the question at point:
Earth's mass: me=5,97e24 kg
Earth's radius: R=6,38e6 m
Moment of inertia: K=(2/5)*Me*R^2=9,72e37 kgm^2 AeB means A*10^B Now we need the moment, which is force F multiplied by R: M=F*R (note: M=/=m_e)
BUT: in relation to Newton's 2nd: M=dL/dt, in which dL=change of angular momentum L; dt=change of time. Angular momentum L=w*K, in which w=2*pi*f and f is Earth's spinning frequency (once every 24h)> w=pi/43200 1/s
We get F*R=dL/dt > F=dL/(r*dt) dL=9,72e37*(pi/42300) 1/s =7,07e33 We can forget the minus sign since we aren't calculating vectors. Let's make an assumption: we want to stop Earth in exactly 1 second > dt=1s > F=7,07e33/6,78e6=1,043e27 Newtons for 1 second at the equator stops the Earth's spinning.
Physics undergrad out.
This problem requires rotational mechanics, i.e. mass is replaced with moment of inertia, force by moment and momentum but angular momentum. Besides, stopping linear movement by applying a force also requires time.
But to the question at point:
Earth's mass: me=5,97e24 kg
Earth's radius: R=6,38e6 m
Moment of inertia: K=(2/5)*Me*R^2=9,72e37 kgm^2 AeB means A*10^B Now we need the moment, which is force F multiplied by R: M=F*R (note: M=/=m_e)
BUT: in relation to Newton's 2nd: M=dL/dt, in which dL=change of angular momentum L; dt=change of time. Angular momentum L=w*K, in which w=2*pi*f and f is Earth's spinning frequency (once every 24h)> w=pi/43200 1/s
We get F*R=dL/dt > F=dL/(r*dt) dL=9,72e37*(pi/42300) 1/s =7,07e33 We can forget the minus sign since we aren't calculating vectors. Let's make an assumption: we want to stop Earth in exactly 1 second > dt=1s > F=7,07e33/6,78e6=1,043e27 Newtons for 1 second at the equator stops the Earth's spinning.
Physics undergrad out.
#103 to #91

McCockAFucker
Reply +1 123456789123345869
(07/05/2013) [] the earths mass is 5.97219 × 10^24 kilograms. and you need to have the centripetal force of the earth or some **** (got it off the internet so dont think im smart enough to figure this out) multiplying it would result in the force required to stop it from spinning being 9.9577042267626e+21kg
#209 to #93

ghettograndpa
Reply 0 123456789123345869
(07/05/2013) [] Wasn't the answer to all things 42?
#214 to #212

ghettograndpa
Reply 1 123456789123345869
(07/05/2013) [] Jesus and the Six Guardians of the gate of Hell