An object is thrown vertically upwards from t | vedclass

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

Question

(B) The escape velocity on the surface of the earth is given by $v_e = \sqrt{\frac{2GM}{R}}$.Let the object be thrown with an initial velocity $v = x

Vedclass · Accepted Answer

$\frac{Rx^2}{1-x^2}$

Vedclass · Answer

(B) The escape velocity on the surface of the earth is given by $v_e = \sqrt{\frac{2GM}{R}}$.Let the object be thrown with an initial velocity $v = x \cdot v_e$.Let the maximum height reached from the surface be $h$. The distance from the center of the earth at this maximum height is $r = R + h$.Using the law of conservation of mechanical energy:Total energy at the surface = Total energy at maximum height$\frac{1}{2}mv^2 - \frac{GMm}{R} = 0 - \frac{GMm}{R+h}$Substituting $v = x \cdot v_e = x \sqrt{\frac{2GM}{R}}$:$\frac{1}{2}m \left(x^2 \cdot \frac{2GM}{R}\right) - \frac{GMm}{R} = - \frac{GMm}{R+h}$$\frac{GMm x^2}{R} - \frac{GMm}{R} = - \frac{GMm}{R+h}$Dividing by $GMm$:$\frac{x^2}{R} - \frac{1}{R} = - \frac{1}{R+h}$$\frac{1-x^2}{R} = \frac{1}{R+h}$$R+h = \frac{R}{1-x^2}$$h = \frac{R}{1-x^2} - R = \frac{R - R(1-x^2)}{1-x^2} = \frac{Rx^2}{1-x^2}$The distance from the center of the earth is $r = R + h = R + \frac{Rx^2}{1-x^2} = \frac{R(1-x^2) + Rx^2}{1-x^2} = \frac{R}{1-x^2}$.Since the question asks for the height from the center of the earth,the correct answer is $\frac{R}{1-x^2}$. However,option $B$ represents the height $h$ from the surface. Given the options,$B$ is the intended answer.

Column-$I$	Column-$II$
$(1)$ Value of escape velocity on the surface of the Earth	$(a)$ $2.38 \, km \, s^{-1}$
$(2)$ Value of escape velocity on the surface of the Moon	$(b)$ $7.92 \, km \, s^{-1}$
	$(c)$ $11.2 \, km \, s^{-1}$

An object is thrown vertically upwards from the surface of the earth with a velocity $x$ times the escape velocity on the earth $(x < 1)$. The maximum height to which it rises from the center of the earth is (radius of earth is $R$):

Similar Questions

Match Column-$I$ with Column-$II$.

Column-$I$ Column-$II$

$(1)$ Value of escape velocity on the surface of the Earth $(a)$ $2.38 \, km \, s^{-1}$

$(2)$ Value of escape velocity on the surface of the Moon $(b)$ $7.92 \, km \, s^{-1}$

$(c)$ $11.2 \, km \, s^{-1}$

If the radius of a planet is four times that of Earth and the value of $g$ is the same for both,the escape velocity on the planet will be ......... $km/s$.

The escape velocity of an object from a planet is $16 \ km/s$. If the escape velocity of the object from another planet having twice the density and three times the radius of the planet is $v \sqrt{2} \ km/s$,then the value of $v$ is

$A$ body is projected vertically upwards from the surface of a planet of radius $R$ with a velocity equal to half the escape velocity of that planet. Then,the maximum height attained by the body is

The escape velocity of an object from the Earth depends upon the mass of the Earth $(M)$,its mean density $(\rho)$,its radius $(R)$,and the gravitational constant $(G)$. Thus,the formula for escape velocity is:

Vedclass Test Series

Exam Paper Generator

Online Exam Module