# Answer to Question #25471 in Mechanics | Relativity for ruru

Question #25471

Help me solve this physics homework:

in sense we all have kinetic energy,even when we are standing still.the earth with a radius of 6.37*10 power 6 m,rotates about its axis once a day.ignoring the earths rotation about the sun,what is the K.E of a 50Kg man standing on the surface of the earth.

in sense we all have kinetic energy,even when we are standing still.the earth with a radius of 6.37*10 power 6 m,rotates about its axis once a day.ignoring the earths rotation about the sun,what is the K.E of a 50Kg man standing on the surface of the earth.

Expert's answer

For the kinetic energy we have the following formula:

T = m*v^2/2

T - kinetic energy

m - mass

v - velocity

To determine the velocity we apply this one (as the Earth is rotating):

v = 2*pi*R/T

R - radius of the Earth

T - period of rotation ( 1 day )

pi = 3.14

v = 2*3.14*6.37*10^6 m / (24*3600 sec ) = 463 m/s

T = m*v^2/2 = 50*463^2/2 J = 5359225 J = 5.36 MJ

T = m*v^2/2

T - kinetic energy

m - mass

v - velocity

To determine the velocity we apply this one (as the Earth is rotating):

v = 2*pi*R/T

R - radius of the Earth

T - period of rotation ( 1 day )

pi = 3.14

v = 2*3.14*6.37*10^6 m / (24*3600 sec ) = 463 m/s

T = m*v^2/2 = 50*463^2/2 J = 5359225 J = 5.36 MJ

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