Question

to
A body is doopped from a a height equal
radius
of the earth. The velocity
before touching the
ground is
acquired by it​

Answer 1

Answer:

to

A body is doopped from a a height equal

radius

of the earth. The velocity

before touching the

ground is

acquired by it

Answer 2

Answer:

v = √gR

Explanation:

Let v be the velocity of the body before touching the ground

m - mass of the body

M - mass of the earth

Since the body is dropped, initial velocity, u = 0

It's dropped from a height equal to the radius of the earth, h = R

Total energy before dropping :

$\sf Kinetic \ Energy,KE_1 = \dfrac{1}{2}mu^2 \\\\ KE_1=\dfrac{1}{2}m(0)^2 \\\\ KE_1=0$

$\sf Potential \ Energy,PE_2 =\dfrac{-GMm}{R+h} \\\\ \sf PE_2=\dfrac{-GMm}{R+R} \\\\ \sf PE_2=\dfrac{-GMm}{2R}$

Total energy before dropping = KE₁ + PE₁

TE₁ = 0 + (-GMm/2R)

$\boxed{\sf TE_1=\dfrac{-GMm}{2R}}$

Total energy when body hits the ground :

$\sf Kinetic \ Energy,KE_2=\dfrac{1}{2}mv^2$

$\sf Potential \ Energy, PE_2=\dfrac{-GMm}{R}$

Total energy when body touches the ground = KE₂ + PE₂

 $\boxed{\sf TE_2=\dfrac{1}{2}mv^2-\dfrac{GMm}{R}}$

Since total energy is conserved.

   TE₁ = TE₂

 $\sf \dfrac{-GMm}{2R}=\dfrac{1}{2}mv^2-\dfrac{GMm}{R} \\\\ \sf \dfrac{1}{2}mv^2=\dfrac{-GMm}{2R}+\dfrac{GMm}{R} \\\\ \dfrac{1}{2}\not{m}v^2=\dfrac{-GM\not{m}}{2R}+\dfrac{GM\not{m}}{R} \\\\ \dfrac{1}{2}v^2=\dfrac{-GM}{2R}+\dfrac{GM}{R} \\\\ \dfrac{1}{2}v^2=\dfrac{-GM+2GM}{2R} \\\\ \dfrac{1}{\not{2}}v^2=\dfrac{GM}{\not{2}R} \\\\ v^2=\dfrac{GM}{R} \\\\ v^2=\dfrac{gR^2}{R} \\\\ v^2=gR \\\\ v=\sqrt{gR}$

The velocity before touching the ground acquired by it​ = √gR