Escape velocity from a planet of mass M and radius R is v. From a planet of mass 4M and radius 2R, escape velocity:
A4v
B2v
Cv
D√2 v
Answer & Solution
Correct answer: D. √2 v
v_esc = √(2GM/R). Original = √(2GM/R). New = √(2G(4M)/(2R)) = √(4GM/R) = √2 × √(2GM/R) = √2 v.
Related questions
Cavendish's torsion balance experiment (1798) measuredThe gravitational field near Earth's surface is approximately uniform becauseNewton showed that the same law that makes an apple fall to the ground alsoFor a body in a circular orbit around Earth of radius r, the ratio of kinetic energy to toThe weight of an object at a certain height above Earth's surface equals its weight at depThe variation of g with depth d below Earth's surface (assuming uniform density) isA geostationary satellite orbits Earth at a height where its period equalsThe gravitational potential energy of two point masses m₁ and m₂ separated by distance r i