Question

A rocket is fired vertically with a speed of 5 km s¯¹ from the earth’s surface. How far from the earth does the rocket go before returning to the earth ? Mass of the earth = 6.0 × 1024 kg; mean radius of the earth = 6.4 × 10⁶ m; G = 6.67 × 10¯¹¹ N m² kg¯².

Answer 1

Given: A rocket is fired vertically with a speed of 5  kms –1 , the mass of the earth is 6.0× 10 24  kg, mean radius of the earth is 6.4× 10 6  m and the value of G is 6.67× 10 −11   Nm 2 /kg.

Total energy of the rocket at the Earth’s surface is given as,

E i = 1 2 m u 2 +( − GMm R )

Where, mis the mass of rocket on the Earth’s surface, u is the speed of rocket, M is the mass of Earth and R is the radius of earth.

At its highest point the potential energy of the rocket is given as,

E f =−( GMm R+h )

Where, h is the height of rocket above the Earth’s surface and v is the speed of rocket at the highest point.

From the law of conservation of energy,

E i = E f

By substituting the given values in the above expression, we get

1 2 m v 2 +( − GMm R )=− GMm R+h 1 2 v 2 =GM( 1 R − 1 R+h ) =GM( R+h−R R( R+h ) ) = GMRh R 2 ( R+h ) (1)

We know that,

g= GM R 2

By substituting the given values in equation (1), we get

1 2 v 2 == gRh ( R+h ) v 2 ( R+h )=2gRh v 2 R=h( 2gR− v 2 ) h= v 2 R ( 2gR− v 2 )

By substituting the given values in the above expression, we get

h= 6.4× 10 6 × ( 5× 10 3 ) 2 2×9.8×6.4× 10 6 − ( 5× 10 3 ) 2 =1.6× 10 6  m

Height achieved by the rocket with respect to the centre of the Earth is given as,

H=R+h

By substituting the given values in the above expression, we get

H=6.4× 10 6 +1.6× 10 6 =8.0× 10 6  m

Thus, the rocket goes 8.0× 10 6  m from the Earth’s centre before returning back.