the escape velocity from the earth surface is 11km/s. a certain planet has a radius twice that of the earth but it's mean density is the same as that of the earth . the value of escape velocity from its planet would be
Answers
Answer:
Explanation:
=> The escape velocity at a given distance:
Ve = √2GM/R ...(1)
=> Now, According to your question,
=> the escape velocity from the earth surface, √2GM/R = 11km/s.
=> a certain planet has a radius twice that of the earth, Rp = 2R
=> it's mean density is the same as that of the earth.
The mass of the given planet (Mp) :
Vp = planet volume = (4/3)*π*Rp³
Ve = earth volume = (4/3)*π*R³
Vp/Ve = Rp³/R³ = (2*R)³/R³ = 2³ = 8
Vp/Ve = 8
Vp = 8Ve
As the density is the same for both planets, Mp = 8*M.
Thus, mass becomes 8 times
=> By putting the values in eq(1), we get
Ve(of planet) = √2GMp/Rp
= √2*G*8M/2R
= √4*2GM/R
= 2* √2GM/R [But √2GM/R = 11 km/s]
= 2 * 11
= 22 km/s
Thus, the value of escape velocity from this planet would be 22km/s.
Answer:
ves=2×11=22
I hope that your answer