Aerospace Engineering > GATE 2019 > Escape Velocity
The product of earth's mass (M) and the universal gravitational constant (G) is GM = 3.986 x 1014 m3/s2. The radius of earth is 6371 km. The minimum increment in the velocity to be imparted to a spacecraft flying in a circular orbit around the earth at an altitude of 4000 km to make it exit earth's gravitational field is _______ km/s (round off to 2 decimal places).
Correct : 2.54 to 2.62
Similar Questions
The universal gravitational constant is 6.67 x 10-11 Nm2/kg2. For a planet of mass 6.4169 x 1023 kg and radius 3390 km, the escape velocity is _______ km/s. (ro...
The product of earth's mass (M) and the universal gravitational constant (G) is GM = 3.986 x 10^14 m^3/s^2. The radius of earth is 6371 km. The minimum incremen...
If → denotes increasing order of intensity, then the meaning of the words dry → arid → parched is analogous to diet → fast → ______. Which one of the given opt...
Total Unique Visitors
Loading......