Question

A particle is released into a deep hole extending to the center of the earth. Determine the velocity of the particle at a depth of one kilometer from the surface. Suppose, g = 10 m / s ° and the radius of the earth is 6400 km​

Answer 1

Answer:

answer for the question is

Explanation:

Correct option is

D

2

v

e

Lets m= mass of the particle

M= mass of the earth

R= radius of the earth

v= speed of particle when it reaches the center of the earth

applying conservation of total energy,

(K.E)

1

+(P.E)

1

=(K.E)

2

+(P.E)

2

0+(−

R

GMm

)=

2

1

mv

2

+(−

2R

3GMm

)

−

R

GM

+

2R

3GM

=

2

1

v

2

v

2

=

R

GM

v=

R

GM

. . . . (1)

We know that the escape velocity of the particle from the earth is given by,

v

e

=

R

2GM

. . . . .(2)

v

e

=

2

R

GM

v

e

=

2

v (from equation 1)

v=

2

v

e

thank you