Determine the escape velocity of a planet with twice the radius but the same density as Earth. Earth's escape velocity is approximately 11.2 km/s.
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The escape velocity is given by the formula:
v = sqrt((2GM)/r)
Where:
v = escape velocity
G = gravitational constant
M = mass of the planet
r = radius of the planet
Since the density of the planet is the same as Earth and the volume of the planet is proportional to the cube of the radius, then the mass of the planet will also be proportional to the cube of the radius.
Let's call the radius of the planet as Rp and the radius of Earth as Re. Since Rp is twice the radius of Earth (2Re), we can say:
Rp = 2 * Re
The mass of the planet, Mp, will be proportional to the cube of the radius:
Mp = (2Re)^3 = 8Re^3
Now we can calculate the escape velocity:
v = sqrt((2GM)/r)
= sqrt((2G(8Re^3))/(2Re))
= sqrt((8GRe^3)/(2Re))
= sqrt(4GRe^2)
= 2sqrt(GRe^2)
Substituting the known values:
v = 2sqrt(GRe^2) = 2 * 11.2 km/s = 22.4 km/s
Therefore, the escape velocity of a planet with twice the radius but the same density as Earth is approximately 22.4 km/s.
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