Answer:
The speed of the object just before it hits Earth is [tex]\sqrt{\dfrac{GM}{R}}[/tex]
(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
[tex]P.E=-\dfrac{GmM}{R+h}[/tex]
The kinetic energy at height above the surface of the earth
[tex]K.E = 0[/tex]
The total energy at height above the surface of the earth
[tex]E = K.E+P.E[/tex]
[tex]E = -\dfrac{GmM}{R+h}[/tex]....(I)
The total energy at the surface of the earth
[tex]E'=\dfrac{1}{2}mv^2-\dfrac{GmM}{R}[/tex]....(II)
We need to calculate the speed of the object just before it hits Earth
From equation (I) and (II)
[tex]-\dfrac{GmM}{R+h}=\dfrac{1}{2}mv^2-\dfrac{GmM}{R}[/tex]
Here, h = R
[tex]-\dfrac{GmM}{2R}=\dfrac{1}{2}mv^2-\dfrac{GmM}{R}[/tex]
[tex]v= \sqrt{\dfrac{GM}{R}}[/tex]
Hence, The speed of the object just before it hits Earth is [tex]\sqrt{\dfrac{GM}{R}}[/tex].
