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Zainab Ansari Zainab Ansari
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A year ago
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?
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Anonymous
wrote...
A year ago
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.
Anonymous
wrote...
A year ago
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.
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