Acceleration due to Gravity; The ratio of escape velocity.

The acceleration due to gravity ( g ) on the surface of the Earth is approximately 2 . 6 times that on the surface of the Mars . Given that the radius of Mars is about one half the radius of the Earth , the ratio of the escape velocity on Earth to that on Mars is approximately . . . . . . . . . . .

a ) 1 . 1
b ) 1 . 3
c ) 2 . 3
d ) 5 . 2 

Solution :

The escape velocity :

The escape velocity can be define as the minimum velocity required to overcome gravitational force of attraction and can escape from the gravitational attraction of any particular planet or other object.

The escape velocity can be represented mathematically as . . . . . .

V _e = ( 2 g R ) ^{1 / 2}

Where : -

V _e          = Escape velocity

g              = Acceleration due to gravity

R             = Radius of the planet.

In the above problem, let us consider

V _{e E}        = The escape velocity on Earth .

V _{e M}       = The escape velocity on Mars .

g _E           = The acceleration due to gravity on Earth .

g _M          =  The acceleration due to gravity on Mars .

R _E           = Radius of the Earth .

R_M          = Radius of the Mars .

The equation for the escape velocity on the earth can be written as

V _{e E}        = ( 2 g _E  R _E ) ^{1 / 2} .. .  . . . . . . . . . . . . . . . . . . . . . . . . . . . . .. . . . . . . . . . . . . . . . . . . . . . . . ( 1 )

Similarly ,

The equation for the escape velocity on the earth can be written as

V _{e M}       = ( 2 g _M  R _M ) ^{1 / 2}.. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . ( 2 )

Therefore the ration of the escape velocity can be given as

V _{e E} /  V _{e M}   =  ( 2 g _E  R _E ) ^{1 / 2}  /  ( 2 g_M  R _M ) ^{1 / 2} 

                                               .. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . ( 3 )

As , The acceleration due to gravity ( g ) on the surface of the Earth is approximately 2 . 6 times that on the surface of the Mars . ( Given ).

Therefore ,

g _E  = 2.6 g _M  . . . . . . . . . . . . . . . . .. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . ( 4 )

and

The radius of Mars is about one half the radius of the Earth,

i . e .

R _E  = 1/2  R _M . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . ( 5 )

Substituting the values of equation ( 4 ) and ( 5 ) in equation ( 3 ) we get

V _{e E} /  V _{e M}     =  ( 2 g _E  R _E ) ^{1 / 2}  /  ( 2 g_M  R _M ) ^{1 / 2}

V _{e E} /  V _{e M}     =  ( 2 [2.6]g _M  [1/2]R _M ) ^{1 / 2}   /  ( 2 g _M  R _M ) ^{1 / 2})

V _{e E} / V _{e M}     =   2 .3

Answer :

the ratio of the escape velocity on Earth to that on Mars is approximately is  2.3  ( c )