find the escape velocity of earth
Answers
Answer:
Explanation:
vc=2GMr−−−−√ ,
where,
vc is the escape velocity
G is the universal gravitational constant
M is the mass of the celestial object whose gravitational pull has to be superseded
r is the distance from the object to the centre of mass of the body to be escaped
From this relation it is obvious that escape velocities for larger planets (or celestial bodies) is greater since it will have a larger mass compared to smaller planets with a lower mass (having less gravity in comparison).
On earth, the escape velocity is around 40,270 kmph, which is around 11,186 m/s.
Answer:
Ve=11.2km
Explanation:
PE+KE=0
where,F=0,g=9.8m,R=6400
W=F×S
-GMm/(R+h)+1/2mve*=0
1/2mve*=GMm/(R+h)
ve*=2GM/R+h
Ve=√2GM/R+h
Rdius of Earth ,
R=6400,h=400
hereR^^^^^h
hence h can be neglected Be=√2GM/R
WKT,
g=GM/R*
GM=gR*
Ve=√2gR*/R
Ve=√2gR
by substituting valuues,
we get,
Ve=11.2km/s