Question

With what horizontal velocity should a body be thrown so that it may move parallel to the surface of the earth along the equator ? The radius of the earth at the equator is 6400km and g=9.7m/s

Answer 1

if object is thrown parallel to the surface of earth and velocity is tangential at equator

then we will have

$F_{net} = F_c$

$mg = \frac{mv^2}{R}$

by solving above equation we will have

$g = \frac{v^2}{R}$

$v^2 = Rg$

$v = \sqrt{Rg}$

$v = \sqrt{6400*10^3 * 9.7}$

$v = 7.88 * 10^3m/s$

so object will move with above speed

Answer 2

Answer: The horizontal velocity of the body $v=7.88\times 10^{ 3 }\frac { m }{ s }$

To assume the motion parallel to the earth surface, the force acting on the body must be equal to the force acting along the radius or center of the earth, which is the centripetal force. Thereby,

                         $F_{ net }={ F }_{ C }$

                         $\Rightarrow mg=\frac { m{ v }^{ 2 } }{ R }$

                         $\Rightarrow g=\frac { { v }^{ 2 } }{ R }$

                         $\Rightarrow { v }^{ 2 }=Rg$

                         $\Rightarrow v=\sqrt { Rg }$

                         $\Rightarrow v=\sqrt { 6400{ \times 10 }^{ 3 }\times 9.7 }$

                         $\Rightarrow v=7.88\times 10^{ 3 }\frac { m }{ s }$