An artificial satellite is moving in a circular orbit around the earth with a speed equal to half of the escape speed from the earth. The height of the satellite from the earth's surface is (given radius of earth = 6400 km)
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___________________❤
_______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
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