A space ship is launched into a circular orbit close to the surface of the earth. The additional velocity now imparted to the spaceship in the orbit to overcome the gravitational pull is

Question

A space ship is launched into a circular orbit close to the surface of the earth. The additional velocity now imparted to the spaceship in the orbit to overcome the gravitational pull is (A) 11.2 km s -1 (B) 8 km s -1 (C) 3.2 km s -1 (D) 1.414 × 8 km s -1

Tardigrade Admin · Accepted Answer

ve=11.2 km / s (m v'2/R)=(G M m/R2) v'2=(G M/R2) R=8 km / s Therefore, the additional velocity =(11.2-8)=3.2 km / s

Q. A space ship is launched into a circular orbit close to the surface of the earth. The additional velocity now imparted to the spaceship in the orbit to overcome the gravitational pull is

$11.2 km s^{- 1}$

$8 km s^{- 1}$

$3.2 km s^{- 1}$

$1.414 \times 8 km s^{- 1}$

$v_{e} = 11.2 km / s$
$\frac{m v ^{^{'} 2}}{R} = \frac{GM m}{R ^{2}}$
$v^{^{'} 2} = \frac{GM}{R ^{2}} R = 8 km / s$
Therefore, the additional velocity $= (11.2 - 8) = 3.2 km / s$

Q. A space ship is launched into a circular orbit close to the surface of the earth. The additional velocity now imparted to the spaceship in the orbit to overcome the gravitational pull is

11.2kms−1

8kms−1

3.2kms−1

1.414×8kms−1

ve​=11.2km/s Rmv′2​=R2GMm​ v′2=R2GM​R=8km/s Therefore, the additional velocity =(11.2−8)=3.2km/s

$11.2 km s^{- 1}$

$8 km s^{- 1}$

$3.2 km s^{- 1}$

$1.414 \times 8 km s^{- 1}$

$v_{e} = 11.2 km / s$
$\frac{m v ^{^{'} 2}}{R} = \frac{GM m}{R ^{2}}$
$v^{^{'} 2} = \frac{GM}{R ^{2}} R = 8 km / s$
Therefore, the additional velocity $= (11.2 - 8) = 3.2 km / s$