The speed at the height ([tex]\frac{R_E}{2}[/tex]) will be [tex]\rm v = \sqrt{\frac{GM}{1.5R_E }[/tex]. While the escapes speed will be [tex]\rm v=\sqrt{\frac{2GM}{R} }[/tex].
What is escape speed?
Escape speed is the minimum speed required for a free, non-propelled object to escape from the gravitational pull of the main body and reach an infinite distance from it in celestial physics.
It is commonly expressed as an ideal speed, neglecting atmospheric friction.
The speed at the height ([tex]\frac{R_E}{2}[/tex]) will be;
Centripetal force is balanced by the gravitational force then,
Centripetal force = gravitaional force
[tex]\rm m\frac{v^2}{r} = \frac{GMm}{r^2} \\\\ \rm v = \sqrt{\frac{GM}{r} } \\\\ \rm v = \sqrt{\frac{GM}{R_E+0.5R_E }[/tex]
[tex]\rm v = \sqrt{\frac{GM}{1.5R_E }[/tex]
Hence the speed at the height ([tex]\frac{R_E}{2}[/tex]) will be [tex]\rm v = \sqrt{\frac{GM}{1.5R_E }[/tex].
The formula for escape speed of the earth will be ;
As we know that energy always gets conserved so potential energy is equal to kinetic energy.
PE = KE
[tex]\rm \frac{GmM}{R^2} = \frac{1}{2} mv^2\\\\ \rm v=\sqrt{\frac{2GM}{R} }[/tex]
Hence the escape speed will be [tex]\rm v=\sqrt{\frac{2GM}{R} }[/tex].
To learn more about the escape speed refer to the link;
https://brainly.com/question/14178880
