# A satellite is moving with constant speed v in a circular orbit around the earth.an object of mass m is ejected from the gravitational pull of the earth.at the time of ejection,the kinetic energy of object is 1) 1/2 mv^2. 2)mv^2 3)3/2 mv^2. 4)2mv^2

2
by kaushikravikant

2016-03-26T22:42:30+05:30

### This Is a Certified Answer

Certified answers contain reliable, trustworthy information vouched for by a hand-picked team of experts. Brainly has millions of high quality answers, all of them carefully moderated by our most trusted community members, but certified answers are the finest of the finest.
EXPLANATION:
For a spherically symmetric body (earth), the escape velocity at a given distance is calculated by the formula Ve = √(2GM/R)

we need the orbital height, which we can get from:
Satellite motion, circular
V = √(GM/R)
V = velocity in m/s
G = 6.673e-11 Nm²/kg²
M is mass of central body in kg
R is radius of orbit in m

V² = GM/R
R = GM/V²

plugging that into the escape velocity equation, we get
Ve = √(2GM/(GM/V²)) = V√2

Kinetic Energy in J if m is in kg and v is in m/s
KE = ½mv² = ½m(V√2)² = mV²

hope it helps..
plss mark it as brainliest.....
Nice ans
2016-03-27T17:41:27+05:30
The correct answers is option 2