The escape velocity at the surface of the Earth is approximately 8 km/s. What is the escape velocity for a planet whose radius is 16 times larger than Earth's, and whose mass is 144 times that of Earth's? 1. 288 km/s 2. 72 km/s 3. 6 km/s 4. 8 km/s 5. 24 km/s
Added by Luis Miguel R.
Step 1
Step 1: Calculate the ratio of the escape speed at the planet to the escape speed at Earth using the formula: \[ \frac{v_{e,planet}}{v_{e,earth}} = \sqrt{\frac{2 \cdot G \cdot M_{planet}}{R_{planet}} \div \frac{2 \cdot G \cdot M_{earth}}{R_{earth}}} \] Show more…
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The escape velocity on earth is 11.2 km/s. On another planet having twice radius and 8 times mass of the earth, the escape velocity will be
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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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