Title: AP Physics Potential EnergyPost by: Zainab Ansari on Feb 1, 2023
3. Planet X-39 has a mass equal to 1/3 that of Earth and a radius equal to 1/3 that of Earth. If v is the escape speed for Earth, what is the escape speed for X-39?
Title: Re: AP Physics Potential EnergyPost by: bio_man on Feb 1, 2023
Escape velocity is the minimum velocity needed for an object to overcome the gravitational force exerted by the planet. It is given by the following relation :
V = √(2GM/r) where, V - escape velocity G - gravitational constant M - mass of planet r - radius of planet We can take the ratio of escape velocities as follow : V1/V2 = √(M1/M2) * √(r2/r1) where, 1 is for Earth and 2 is for X-39 given that, V1 = V M2 = (1/3)*M1 => M1/M2 = 3 r2 = (1/3)*r1 => r2/r1 = 1/3 therefore, V/V2 = √3 * √(1/3) = 1 ie. V2 = V ie. Escape velocity of X-39 is same as that of Earth. Title: Re: AP Physics Potential EnergyPost by: bio_man on Feb 1, 2023
The escape speed for a planet is determined by its mass and radius, and it can be calculated using the formula:
v = √(2GM/R) where G is the gravitational constant, M is the mass of the planet, and R is its radius. Since X-39 has a mass equal to 1/3 that of Earth and a radius equal to 1/3 that of Earth, the escape speed for X-39 can be calculated as follows: v = √(2G * (1/3M) / (1/3R)) = √(2G * M / R) / √(1/3) = √(2GM/R) / √(1/3) = v / √(1/3) So the escape speed for X-39 is 1/√(3) times the escape speed for Earth. |