What is the escape velocity (in km/s) from a planet of mass 11.7 times the mass of the earth and has the same radius as the earth? The escape velocity from the earth is 11.2 km/s.
Added by John B.
Step 1
2 km/s, we can use the formula for escape velocity: \[v = \sqrt{\frac{2GM}{r}}\] where: - \(v\) is the escape velocity, - \(G\) is the gravitational constant, - \(M\) is the mass of the planet, and - \(r\) is the radius of the planet. Show more…
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The escape speed from the surface of the Earth is $11.2 \mathrm{km} / \mathrm{s} .$ What would be the escape speed from another planet of the same density (mass per unit volume) as Earth but with a radius twice that of Earth?
For the earth escape velocity is $11.2 \mathrm{~km} / \mathrm{s}$. What will be the escape velocity of that planet whose mass and radius are four times those of earth? (a) $11.2 \mathrm{~km} / \mathrm{s}$ (b) $44.8 \mathrm{~km} / \mathrm{s}$ (c) $2.8 \mathrm{~km} / \mathrm{s}$ (d) $0.7 \mathrm{~km} / \mathrm{s}$
Escape velocity is the minimum speed that an object must reach to escape the pull of a planet's gravity. Escape velocity $v$ is given by the equation $v=\sqrt{\frac{2 G m}{r}},$ where $m$ is the mass of the planet, $r$ is its radius, and $G$ is the universal gravitational constant, which has a value of $G=6.67 \times 10^{-11} \mathrm{m}^{3} / \mathrm{kg} \cdot \mathrm{s}^{2} .$ The mass of Earth is $5.97 \times 10^{24} \mathrm{kg},$ and its radius is $6.37 \times 10^{6} \mathrm{m} .$ Use this information to find the escape velocity for Earth in meters per second. Round to the nearest whole number. (Source: National Space Science Data Center)
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