Question
What's the total mechanical energy of a $500-\mathrm{kg}$ satellite in circular orbit $1500 \mathrm{~km}$ above Earth's surface?
Step 1
We do this by multiplying the given value by 1000 (since 1 km = 1000 m). So, $h = 1500 \times 1000 = 1.5 \times 10^6$ m. Show more…
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A 500 -kg satellite is in a circular orbit at an altitude of $500 \mathrm{~km}$ above the Earth's surface. Because of air friction, the satellite is eventually brought to the Earth's surface, and it hits the Earth with a speed of $2.00 \mathrm{~km} / \mathrm{s}$. How much energy was transformed to internal energy by means of friction?
A 500 -kg satellite is in a circular orbit at an altitude of $500 \mathrm{km}$ above the Earth's surface. Because of air friction, the satellite eventually falls to the Earth's surface, where it hits the ground with a speed of $2.00 \mathrm{km} / \mathrm{s}$. How much energy was transformed into internal energy by means of air friction?
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