A satellite is moving in a circular orbit around the earth with a speed equal to half the escape speed from the earth. If 'R' is the radius of the earth, then the height of the satellite above the surface of the earth is ..............please help me
Answers
Answered by
0
The height of the satellite above the surface of the earth is 2R
harshitha05:
but answer is R..... I want explanation
Answered by
1
Hello
___________________❤
_______Answer ⤵⤵⤵
V = √(GM/R)
V = velocity in m/s
G = 6.673e-11 Nm²/kg²
V² = GM/R
R = GM/V²
We get,
Ve = √(2GM/(GM/V²) = V√2
Kinetic Energy = KE = ½mv² = ½m(V√2)² = mV²
___________________❤
_______Answer ⤵⤵⤵
V = √(GM/R)
V = velocity in m/s
G = 6.673e-11 Nm²/kg²
V² = GM/R
R = GM/V²
We get,
Ve = √(2GM/(GM/V²) = V√2
Kinetic Energy = KE = ½mv² = ½m(V√2)² = mV²
Similar questions