Question

An object is dropped from an altitude of one Earth radiusabove Earth's surface. If M is the mass of Earth and R is itsradius the speed of the object just before it hits Earth isgiven by: A. √GM/R B. √GM/2R C. √2GM/R D. √GM/R2 E. √GM/2R2

Answer 1

Answer:

The speed of the object just before it hits Earth is  $\sqrt{\dfrac{GM}{R}}$

(A) is correct option.

Explanation:

Given that,

M = mass of earth

R = radius of earth

The potential energy at height above the surface of the earth

$P.E=-\dfrac{GmM}{R+h}$

The kinetic energy at height above the surface of the earth

$K.E = 0$

The total energy at height above the surface of the earth

$E = K.E+P.E$

$E = -\dfrac{GmM}{R+h}$....(I)

The total energy at the surface of the earth

$E'=\dfrac{1}{2}mv^2-\dfrac{GmM}{R}$....(II)

We need to calculate the speed of the object  just before it hits Earth

From equation (I) and (II)

$-\dfrac{GmM}{R+h}=\dfrac{1}{2}mv^2-\dfrac{GmM}{R}$

Here, h = R

$-\dfrac{GmM}{2R}=\dfrac{1}{2}mv^2-\dfrac{GmM}{R}$

$v= \sqrt{\dfrac{GM}{R}}$

Hence, The speed of the object just before it hits Earth is  $\sqrt{\dfrac{GM}{R}}$.