The escape speed of a projectile on the earth | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(B) The escape velocity of a projectile from the Earth is $v_{esc} = 11.2 \; km/s$.The projection velocity of the projectile is $v_p = 3 v_{esc}$.Let

Vedclass · Accepted Answer

$31.68$

Vedclass · Answer

(B) The escape velocity of a projectile from the Earth is $v_{esc} = 11.2 \; km/s$.The projection velocity of the projectile is $v_p = 3 v_{esc}$.Let the mass of the projectile be $m$.According to the law of conservation of energy,the total energy at the Earth's surface equals the total energy far away from the Earth (where gravitational potential energy is zero).$\frac{1}{2} m v_p^2 - \frac{G M m}{R} = \frac{1}{2} m v_f^2 + 0$Since the escape velocity is defined as $v_{esc} = \sqrt{\frac{2GM}{R}}$,we have $\frac{GM}{R} = \frac{v_{esc}^2}{2}$.Substituting this into the energy equation: $\frac{1}{2} m v_p^2 - \frac{1}{2} m v_{esc}^2 = \frac{1}{2} m v_f^2$.$v_f = \sqrt{v_p^2 - v_{esc}^2} = \sqrt{(3 v_{esc})^2 - v_{esc}^2} = \sqrt{8 v_{esc}^2} = v_{esc} \sqrt{8}$.$v_f = 11.2 	imes 2.828 = 31.68 \; km/s$.

The escape speed of a projectile on the earth's surface is $11.2 \; km/s$. $A$ body is projected out with thrice this speed. What is the speed (in $km/s$) of the body far away from the earth? Ignore the presence of the sun and other planets.

Similar Questions

The escape velocity for the Earth is $11 \ km/s$. The escape velocity for a planet having twice the radius and the same density as the Earth is .......... $km/s$.

The escape velocity of a sphere of mass $m$ is given by ($G =$ Universal gravitational constant; $M_e =$ Mass of the earth and $R_e =$ Radius of the earth).

Escape velocity at the surface of the Earth is $11.2 \, km/s$. If the radius of a planet is double that of the Earth but its mean density is the same as that of the Earth,then the escape velocity will be ........ $km/s$.

$A$ body is projected vertically upward from the surface of the earth with a velocity equal to half the escape velocity. If $R$ is the radius of the earth,the maximum height attained by the body is:

If $v_e$ is escape velocity and $v_0$ is orbital velocity of a satellite for an orbit close to the earth's surface,then these are related by:

Vedclass Test Series

Exam Paper Generator

Online Exam Module