calculate the escape velocity on the surface of a planet given the mass and radius of the planet to be 7.34×10 raise to 22 kg and 1×10raise to 6 m respectively ( gravitional acceleration on the planet = 2m/second square.
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mass = M = 7.34×10^22 kg
radius = R = 1×10^6m
G = gravitational constant = 6.67Nm²/kg²
escape velocity = v = ?
g = GM/R²
___________
v = √2Rg
v = √2RGM/R²
v = √2GM/R
v = √2×6.67×10^-11 × 7.34×10^22/1×10^6
v = √97.92 × 10^-11+22/10^6
v = √97.92×10^11/10^6
v = √97.92×10^11-6
v = √97.92 × 10^5
v = √97.92 × 10¹ × 10^4
v = 10²√979.2 m/s
v = 10²×31.29
v = 3.129 × 10³
so escape velocity is 3.129 × 10³ m/s or 3129 m/s.
radius = R = 1×10^6m
G = gravitational constant = 6.67Nm²/kg²
escape velocity = v = ?
g = GM/R²
___________
v = √2Rg
v = √2RGM/R²
v = √2GM/R
v = √2×6.67×10^-11 × 7.34×10^22/1×10^6
v = √97.92 × 10^-11+22/10^6
v = √97.92×10^11/10^6
v = √97.92×10^11-6
v = √97.92 × 10^5
v = √97.92 × 10¹ × 10^4
v = 10²√979.2 m/s
v = 10²×31.29
v = 3.129 × 10³
so escape velocity is 3.129 × 10³ m/s or 3129 m/s.
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