Question

If a satellite orbits the earth at 2,000 km above sea level, how fast must the
orbiting satellite travel to maintain a circular orbit?​

Answer 1

Answer:

$\huge\purple{Answer}$

Let v be the satellite’s velocity, M the mass of the Earth, r the distance from the Earth’s center (height plus the radius of the Earth), and G Newton’s gravitational constant. Then:

v2r=GMr2 , so v=GMr−−−√

We use the meter-kilogram-second (mks) system of units; in these units:

G=6.67×10−11,M=5.97×1024,GM=3.98×1014 , so:

v=1.99×107r√

We can use this expression to solve for the orbital velocity at any distance r . You’ve specified 2000 km above the surface, and r here is distance from the center of the earth, so for your calculation r is 8378 km. We’re doing this in meters, so r=8.38×106 . We can now solve for v

v=1.99×107r√=v=1.99×1078.38×106√=1.99×1072.89×103=6885 meters per second, or 24,786 km/hour.

Answer 2

Explanation:

Let v be the satellite's velocity, M the mass of the Earth, r the distance from the Earth's center (height plus the radius of the Earth), and G Newton's gravitational constant. Then:

v2r-GMr2, so v=GMr---

We use the meter-kilogram-second (mks) system of units; in these units:

G=6.67×10-11,M=5.97×1024,GM=3.98×1014,

SO:

v=1.99x107rv

We can use this expression to solve for the orbital velocity at any distance r. You've specified 2000 km above the surface, and r here is distance from the center of the earth, so for your calculation r is 8378 km. We're doing this in meters, so r=8.38×106 . We can now solve for v

v=1.99x107r√=v=1.99x1078.38×106√-1.99x107: meters per second, or 24,786 km/hour.