A spaceship goes into a circular orbit close to the earth's surface. What additional velocity must be imparted to the ship so that it is able to escape the gravitational pull of the earth? (R $= 6400$ km, g $= 9.8$ m $s^{- 2}$ )

Question

A spaceship goes into a circular orbit close to the earth's surface. What additional velocity must be imparted to the ship so that it is able to escape the gravitational pull of the earth? (R

= 6400

km, g

= 9.8

m

s^{- 2}

)

Toppr Admin · Accepted Answer

The correct option is B 3-278km s-x2212-1-The orbital velocity in a circular orbit close to the earth is v-x221A-gR-The velocity required to escape ve-x221A-2gR-Hence additional velocity required is ve-x2212-v-x221A-2-x2212-1-x221A-gR-Therefore-ve-x2212-v-0-414-xD7-x221A-9-8-xD7-6400-xD7-103-3278-71m-s -3-278km s-x2212-1-

A spaceship goes into a circular orbit close to the earth's surface. What additional velocity must be imparted to the ship so that it is able to escape the gravitational pull of the earth? (R=6400km, g=9.8 m s−2)7km s−1.3.278km s−1.9km s−1.5km s−1.

The correct option is B 3.278km s−1.The orbital velocity in a circular orbit close to the earth is v=√gR.The velocity required to escape ve=√2gR.Hence additional velocity required is ve−v=(√2−1)√gR.Therefore,ve−v=0.414×√9.8×6400×103=3278.71m/s =3.278km s−1.