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Gravitation

270472 A body is dropped from a height equal to radius of the earth. The velocity acquired by it before touching the ground is

1

V = \sqrt{2 g R}

2

V = 3 gR

3

V = \sqrt{g R}

4

V = 2 gR

Gravitation

270473 When projectile attains escape velocity, then on the surface of planet, its

1

K E < P E

2

P E < K E

3

K E = P E

4

K E = 2 P E

Gravitation

270474 A satellite is moving with constant speed ' $V$ ' in a circular orbit around earth. The kinetic energy of the satellite is

1

\frac{1}{2} m V^{2}

2

m V^{2}

3

\frac{3}{2} m V^{2}

4

2 m V^{2}

Gravitation

270528 A satellite moves around the earth in a circular orbit with speed ' $v$ '. If ' $m$ ' is mass of the satellite then its total energy is

1

\frac{1}{2} {mv}^{2}

2

{mv}^{2}

3

- \frac{1}{2} m v^{2}

4

\frac{3}{2} m v^{2}

Explanation:

TE $= - KE = - \frac{1}{2} m v^{2}$

Gravitation

270529 The K.E. of a satellite in an orbit close to the surface of the earth is E. Its max K.E. so as to escape from the gravitational field of the earth is.

1

2 E

2

4 E

3

2 \sqrt{2} E

4

\sqrt{2} E

Explanation:

$\frac{K_{e}}{K_{0}} = \frac{2 g R}{g R} \Rightarrow K_{e} = 2 K_{0}$

Gravitation

270472 A body is dropped from a height equal to radius of the earth. The velocity acquired by it before touching the ground is

1

V = \sqrt{2 g R}

2

V = 3 gR

3

V = \sqrt{g R}

4

V = 2 gR

Gravitation

270473 When projectile attains escape velocity, then on the surface of planet, its

1

K E < P E

2

P E < K E

3

K E = P E

4

K E = 2 P E

Gravitation

270474 A satellite is moving with constant speed ' $V$ ' in a circular orbit around earth. The kinetic energy of the satellite is

1

\frac{1}{2} m V^{2}

2

m V^{2}

3

\frac{3}{2} m V^{2}

4

2 m V^{2}

Gravitation

270528 A satellite moves around the earth in a circular orbit with speed ' $v$ '. If ' $m$ ' is mass of the satellite then its total energy is

1

\frac{1}{2} {mv}^{2}

2

{mv}^{2}

3

- \frac{1}{2} m v^{2}

4

\frac{3}{2} m v^{2}

Explanation:

TE $= - KE = - \frac{1}{2} m v^{2}$

Gravitation

270529 The K.E. of a satellite in an orbit close to the surface of the earth is E. Its max K.E. so as to escape from the gravitational field of the earth is.

1

2 E

2

4 E

3

2 \sqrt{2} E

4

\sqrt{2} E

Explanation:

$\frac{K_{e}}{K_{0}} = \frac{2 g R}{g R} \Rightarrow K_{e} = 2 K_{0}$

Gravitation

270472 A body is dropped from a height equal to radius of the earth. The velocity acquired by it before touching the ground is

1

V = \sqrt{2 g R}

2

V = 3 gR

3

V = \sqrt{g R}

4

V = 2 gR

Gravitation

270473 When projectile attains escape velocity, then on the surface of planet, its

1

K E < P E

2

P E < K E

3

K E = P E

4

K E = 2 P E

Gravitation

270474 A satellite is moving with constant speed ' $V$ ' in a circular orbit around earth. The kinetic energy of the satellite is

1

\frac{1}{2} m V^{2}

2

m V^{2}

3

\frac{3}{2} m V^{2}

4

2 m V^{2}

Gravitation

270528 A satellite moves around the earth in a circular orbit with speed ' $v$ '. If ' $m$ ' is mass of the satellite then its total energy is

1

\frac{1}{2} {mv}^{2}

2

{mv}^{2}

3

- \frac{1}{2} m v^{2}

4

\frac{3}{2} m v^{2}

Explanation:

TE $= - KE = - \frac{1}{2} m v^{2}$

Gravitation

270529 The K.E. of a satellite in an orbit close to the surface of the earth is E. Its max K.E. so as to escape from the gravitational field of the earth is.

1

2 E

2

4 E

3

2 \sqrt{2} E

4

\sqrt{2} E

Explanation:

$\frac{K_{e}}{K_{0}} = \frac{2 g R}{g R} \Rightarrow K_{e} = 2 K_{0}$

Gravitation

270472 A body is dropped from a height equal to radius of the earth. The velocity acquired by it before touching the ground is

1

V = \sqrt{2 g R}

2

V = 3 gR

3

V = \sqrt{g R}

4

V = 2 gR

Gravitation

270473 When projectile attains escape velocity, then on the surface of planet, its

1

K E < P E

2

P E < K E

3

K E = P E

4

K E = 2 P E

Gravitation

270474 A satellite is moving with constant speed ' $V$ ' in a circular orbit around earth. The kinetic energy of the satellite is

1

\frac{1}{2} m V^{2}

2

m V^{2}

3

\frac{3}{2} m V^{2}

4

2 m V^{2}

Gravitation

270528 A satellite moves around the earth in a circular orbit with speed ' $v$ '. If ' $m$ ' is mass of the satellite then its total energy is

1

\frac{1}{2} {mv}^{2}

2

{mv}^{2}

3

- \frac{1}{2} m v^{2}

4

\frac{3}{2} m v^{2}

Explanation:

TE $= - KE = - \frac{1}{2} m v^{2}$

Gravitation

270529 The K.E. of a satellite in an orbit close to the surface of the earth is E. Its max K.E. so as to escape from the gravitational field of the earth is.

1

2 E

2

4 E

3

2 \sqrt{2} E

4

\sqrt{2} E

Explanation:

$\frac{K_{e}}{K_{0}} = \frac{2 g R}{g R} \Rightarrow K_{e} = 2 K_{0}$

Gravitation

270472 A body is dropped from a height equal to radius of the earth. The velocity acquired by it before touching the ground is

1

V = \sqrt{2 g R}

2

V = 3 gR

3

V = \sqrt{g R}

4

V = 2 gR

Gravitation

270473 When projectile attains escape velocity, then on the surface of planet, its

1

K E < P E

2

P E < K E

3

K E = P E

4

K E = 2 P E

Gravitation

270474 A satellite is moving with constant speed ' $V$ ' in a circular orbit around earth. The kinetic energy of the satellite is

1

\frac{1}{2} m V^{2}

2

m V^{2}

3

\frac{3}{2} m V^{2}

4

2 m V^{2}

Gravitation

270528 A satellite moves around the earth in a circular orbit with speed ' $v$ '. If ' $m$ ' is mass of the satellite then its total energy is

1

\frac{1}{2} {mv}^{2}

2

{mv}^{2}

3

- \frac{1}{2} m v^{2}

4

\frac{3}{2} m v^{2}

Explanation:

TE $= - KE = - \frac{1}{2} m v^{2}$

Gravitation

270529 The K.E. of a satellite in an orbit close to the surface of the earth is E. Its max K.E. so as to escape from the gravitational field of the earth is.

1

2 E

2

4 E

3

2 \sqrt{2} E

4

\sqrt{2} E

Explanation:

$\frac{K_{e}}{K_{0}} = \frac{2 g R}{g R} \Rightarrow K_{e} = 2 K_{0}$