Two bodies, each of mass $M,$ are kept fixed at a separation $2 L$ . A particle of mass m is projected from the midpoint of the line joining their centres, perpendicular to the line. The gravitational constant is $G .$ The correct statement(s) is (are)

Question

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Answer 1

$\frac{- 2 G M m}{L} + \frac{1}{2} m v^{2} = 0$

$\Rightarrow v = 2 \frac{G M}{L}$

Note:
The energy of mass m means its kinetic energy (KE) only and not the potential energy of interaction between m and the two bodies (of mass M each) which is the potential energy of the system.

Answer 2

\frac{- 2 G M m}{L} + \frac{1}{2} m v^{2} = 0

\Rightarrow v = 2 \frac{G M}{L}

Note:
The energy of mass m means its kinetic energy (KE) only and not the potential energy of interaction between m and the two bodies (of mass M each) which is the potential energy of the system.

Two bodies, each of mass $M,$ are kept fixed at a separation $2 L$ . A particle of mass m is projected from the midpoint of the line joining their centres, perpendicular to the line. The gravitational constant is $G .$ The correct statement(s) is (are)

The minimum initial velocity of the mass m to escape the gravitational field of the two bodies is $4 \frac{G M}{L}$

The minimum initial velocity of the mass m to escape the gravitational field of the two bodies is $2 \frac{G M}{L} .$

The minimum initial velocity of the mass m to escape the gravitational field of the two bodies is $\frac{2 G M}{L}$

The energy of the mass m remains constant.

$\frac{- 2 G M m}{L} + \frac{1}{2} m v^{2} = 0$

$\Rightarrow v = 2 \frac{G M}{L}$

Note:
The energy of mass m means its kinetic energy (KE) only and not the potential energy of interaction between m and the two bodies (of mass M each) which is the potential energy of the system.

If 'd' is the distance between the centers of the earth of mass $M_{1}$ and moon of mass $M_{2}$ , then the velocity with which a body should be projected from the mid point of the line joining the earth and the moon, so that it just escapes is :

Masses and radii of earth and moon are $M_{1}, M_{2}$ and $R_{1}, R_{2}$ respectively. The distance between theircentre is 'd'. The minimum velocity given to mass 'M' from the mid point of line joining their centre so that it will escape.

The masses and radii of the earth and moon are $M_{1}$ , $R_{1}$ and $M_{2}$ , $R_{2}$ respectively. Their centres are at distance $d$ apart. The minimum velocity with which a particle of mass $m$ should be projected from a point midway between their centres so that it escapes to infinity is:

Two bodies, each of mass M, are kept fixed at a separation 2L. A particle of mass m is projected from the midpoint of the line joining their centres, perpendicular to the line. The gravitational constant is G.The correct statement(s) is (are)

The minimum initial velocity of the mass m to escape the gravitational field of the two bodies is 4LGM​​

The minimum initial velocity of the mass m to escape the gravitational field of the two bodies is 2LGM​​.

The minimum initial velocity of the mass m to escape the gravitational field of the two bodies isL2GM​​

The energy of the mass m remains constant.

L−2GMm​+21​mv2=0 ⇒v=2LGM​​ Note: The energy of mass m means its kinetic energy (KE) only and not the potential energy of interaction between m and the two bodies (of mass M each) which is the potential energy of the system.

If 'd' is the distance between the centers of the earth of mass M1​ and moon of mass M2​, then the velocity with which a body should be projected from the mid point of the line joining the earth and the moon, so that it just escapes is :

Masses and radii of earth and moon are M1​,M2​ and R1​,R2​ respectively. The distance between theircentre is 'd'. The minimum velocity given to mass 'M' from the mid point of line joining their centre so that it will escape.

The masses and radii of the earth and moon are M1​ , R1​ and M2​, R2​ respectively. Their centres are at distance d apart. The minimum velocity with which a particle of mass m should be projected from a point midway between their centres so that it escapes to infinity is:

Two bodies, each of mass $M,$ are kept fixed at a separation $2 L$ . A particle of mass m is projected from the midpoint of the line joining their centres, perpendicular to the line. The gravitational constant is $G .$ The correct statement(s) is (are)

The minimum initial velocity of the mass m to escape the gravitational field of the two bodies is $4 \frac{G M}{L}$

The minimum initial velocity of the mass m to escape the gravitational field of the two bodies is $2 \frac{G M}{L} .$

The minimum initial velocity of the mass m to escape the gravitational field of the two bodies is $\frac{2 G M}{L}$

$\frac{- 2 G M m}{L} + \frac{1}{2} m v^{2} = 0$

$\Rightarrow v = 2 \frac{G M}{L}$

Note:
The energy of mass m means its kinetic energy (KE) only and not the potential energy of interaction between m and the two bodies (of mass M each) which is the potential energy of the system.

If 'd' is the distance between the centers of the earth of mass $M_{1}$ and moon of mass $M_{2}$ , then the velocity with which a body should be projected from the mid point of the line joining the earth and the moon, so that it just escapes is :

Masses and radii of earth and moon are $M_{1}, M_{2}$ and $R_{1}, R_{2}$ respectively. The distance between theircentre is 'd'. The minimum velocity given to mass 'M' from the mid point of line joining their centre so that it will escape.

The masses and radii of the earth and moon are $M_{1}$ , $R_{1}$ and $M_{2}$ , $R_{2}$ respectively. Their centres are at distance $d$ apart. The minimum velocity with which a particle of mass $m$ should be projected from a point midway between their centres so that it escapes to infinity is: