Recent content by kaminasimon

Sorry i mean the velocity we supplied is smaller if we launched it on equator and it depends on the angle we launched. Sorry i forget it is the velocity we supplied and THE ESCAPE SPEED DOES NOT DEPENDS ON REFERENCE FRAME AND THE ANGLE WE LAUNCHED.thanks i am understand now. The escape speed is...

I think it depends on the angle and absolutely the mass escapes the gravitational field of earth. If the angle is enough small to escape the field the the velocity we supplied will be smaller. Because Vector(V)= vector(v1)+vector(v2). The module V^2=(v1)^2+(v2)^2+2(v1)(v2)cos(c) where c is angle...

Thank you very much

Thanks for answer

I know. The escape velocity depends on the angle we launched. I forgot tell you i assumming the angle is 90 (perpendicular to surface)

We knew on the surface v= (2π÷T)×R which R is radius of Earth and T is period and we have V^2>=(GM÷R)-[(4π^2)÷T^2]×R^2 and result is Min V= 7.893961 km/s

Guy, i have a problem. When we use conservation energy to find the escape velocity, ie the root of potentiality is unlimited, why we don't compute the velocity of earth? We knew the Earth rotated its axis, then everything in surface was had linear velocity. So when we launch a thing, it's...