Question

A body is projected away from earth surface with a speed thrice its escape velocity what will be the speed of body at infinity if it’s escape velocity is 15 km/sec

Answer 1

Answer:

A body is projected away from earth surface with a speed thrice its escape velocity what will be the speed of body at infinity if it’s escape velocity is 15 km/secThe escape velocity of a projectile on the earth surface is 11.2km/sec. A body is projected out with thrice this speed what will be the speed of the body far away from earth that is at infinity

Explanation:

A body is projected away from earth surface with a speed thrice its escape velocity what will be the speed of body at infinity if it’s escape velocity is 15 km/sec

The escape velocity of a projectile on the earth surface is 11.2km/sec. A body is projected out with thrice this speed what will be the speed of the body far away from earth that is at infinitybody attains a height equal to the radius of the earth when projected from earth surface. The velocity of the body with which it was projected

Answer 2

Answer:

The correct answer will be -

The velocity of the body at infinity will be 42.42 km/sec.

Explanation:

Given in the question -

• A body is projected away from the earth's surface.
• With a speed thrice it's escape velocity. Let escape velocity be $v_e$ then the velocity of the projected body $v_p$ will be $v_p = 3v_e$.
• The escape velocity $v_e$ is 15 km/sec. From here we can say that the velocity of the projectile will be = 45 km/sec.

Now we have to find the velocity of the object at infinity. Let's call it $v_i$.

On Earth -

Both gravitational potential energy and kinetic energy will be in consideration.

The energy will be = $\frac{1}{2}mv_p^2 -\frac{1}{2} mv_e^2$

Here potential energy is written in relation to $v_e$.

At infinity -

At infinity, the gravitational potential energy will be 0.

The energy will be = $\frac{1}{2} mv_i^2$

According to the law of conservation of energy, both should be conserved.

$\frac{1}{2} mv_i^2 =\frac{1}{2}mv_p^2 - \frac{1}{2}mv_e^2\\\\v_i^2 = v_p^2 - v_e^2\\\\v_i^2 = (3v_e)^2 - v_e^2\\\\v_i^2 = 8v_e^2\\\\v_i = \sqrt{8} v_e\\ \\v_i = \sqrt{8} \times 15 = 42.42\ km/sec$

So, the velocity of the body at infinity will be 42.42 km/sec.

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