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relation between escape velocity and orbital velocity

ve =21/2vo

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  Let VE be the escape velocity and VO be the orbital velocity, then

VE 2= 2gRE 

VO 2= gRE     ,     where RE is the radius of the earth.

==)  VE 2 = 2 VO 2

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The square of escape velocity is equivalent to twice the square of orbital velocity.

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Vo=under root GM/R+h

if h<<<<R

Vo=under root GM/R

Vo=under root 2GM/R

Ve=under root 2Vo..

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let Ve be the escape velocity..and Vo is the orbital velocity,then

VE2= 2gRE

VO2= gRE , where REis the radius of the earth.

==) VE2= 2 VO2

or

VE = root 2 Vo

this means the square of escape velocity is equivalent to twice the square of orbital velocity.....

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here escape velocity's square =2(orbital velocity)

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Escape velocity2= 2 × Orbital velocity
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Can u explain a relation between orbital velocity ,height ,and time period.
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Escape Velocity = 2^1/2 * Orbital Velocity
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ve= 21/2vo
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ve= 21/2vo
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ve= 21/2vo
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ve^2=2RM/R  WHERE ve=escape speed , R = radius of earth , M = mass of earth , G = universal G constant


Vo^2=GM/R                    WHERE Vo=orbital speed , R = radius of earth , M = mass of earth , G = universal G constant




Ve=(2^1/2 )Vo
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the escape velocity of a body depends upon mass as
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Escape velocity=root 2 Orbital velocity
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Ve=?2Vo

Ve-->Escape velocity
Vo-->Orbital velocity
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Ve=root2 Vo
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Relation between escape velocity and orbital velocity is
V(escape)/V(orbital)=?2
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good
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V escape/V orbital=?2
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escape velocity is the minimum speed needed for an object to "break free" from the gravitational attraction of a massive body. The escape velocity from Earth is about 40,270 km/h (25,020 mph). The velocity of this orbit depends on the distance from the object to the center of the Earth
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1600 km
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Escape velocity is minimum velocity with which a body should be projected so that it just escapes the earth's gravitational field whereas orbital velocity is velocity with which satellite revolve around the earth or the velocity required to put the satellite into its orbit around the earth.
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This is the answer I hope

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Ve=?2gr
Vo =?gr
Ve/Vo = ?2gr /?gr =?2
Ve= ?2vo
Hence escape velocity of a body from the earth's surface is ?2 times it's velocity in a circular orbit just above the earth's surface.
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Ve= ?2V0
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