The velocity with which a projectile must be fired so that it escapes earth's gravitation does not depend on

Mass of the earth

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Mass of the projectile

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Radius of the projectile's orbit

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Gravitational constant

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Solution

The correct option is B Mass of the projectile
Using this we calculate escape velocity:
$v_{e} = \sqrt{\frac{2 G M}{r}}$
In this expression we can see mass of earth(M),Radius of orbit(R), Gravitational constant(G). Mass of projectile does not matter.

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Similar questions

Q. Assertion :The escape speed does not depend on the direction in which the projectile is fired. Reason: Attaining the escape speed is easier if a projectile is fired in the direction the launch site is moving as the earth rotates about its axis.

Q. If

v_{e}

is the escape velocity for earth when a projectile is fired from the surface of earth. Then the escape velocity if the same projectile is fired from its centre is

Q. A projectile is fired vertically from the Earth with a velocity

k v_{e}

where

v_{e}

is the escape velocity and

k

is a constant less than unity. The maximum height to which projectile rises, as measured from the centre of Earth, is

Q. A projectile is fired from the surface of the earth with a velocity of

5 m/s

at an angle

θ

with the horizontal. Another projectile, fired from another planet with a velocity of

3 m/s

at the same angle, follows a trajectory which is identical with the trajectory of the projectile fired from the earth. The value of acceleration due to gravity on the planet (in

m s^{- 2}

) is

({Given g = 9.8 ms}^{- 2})

A projectile rises upto a maximum distance of $\frac{R}{1 - k^{2}}$ where K is a constant and R is the radius of Earth. If the velocity of the projectile with which it should be fired upwards from the surface of Earth to reach this height is equal to the product of a coefficient and escape velocity, then this coefficient is equal to:

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The velocity with which a projectile must be fired so that it escapes earth's gravitation does not depend on

The correct option is B Mass of the projectileUsing this we calculate escape velocity:ve=√2GMrIn this expression we can see mass of earth(M),Radius of orbit(R), Gravitational constant(G). Mass of projectile does not matter.

The correct option is B Mass of the projectile
Using this we calculate escape velocity:
$v_{e} = \sqrt{\frac{2 G M}{r}}$
In this expression we can see mass of earth(M),Radius of orbit(R), Gravitational constant(G). Mass of projectile does not matter.