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42,244,000 meters
- radius = 42,244,000 meters
www.1728.org/kepler3a.htm
People also ask
How do you calculate the orbital radius of a satellite?
How do you find the height of a satellite?
How high should a satellite orbit the Earth?
How far above Earth does a satellite orbit the Moon?
How do you find the orbital radius of the Moon?
How is the period of a satellite related to the distance?
Approximating the figure of Earth by an Earth spheroid (an oblate ellipsoid), the radius ranges from a maximum (equatorial radius, denoted a) of nearly 6,378 km (3,963 mi) to a minimum (polar radius, denoted b) of nearly 6,357 km (3,950 mi).
The orbital speed can be found using v = SQRT(G*M/R). The R value (radius of orbit) is the earth's radius plus the height above the earth - in this case, 6.59 x 10 6 m. Substituting and solving yields a speed of 7780 m/s.
The standard acceleration of gravity for Earth is defined (CODATA 2018) as 9.80665 m/s 2 (exact).
A geostationary orbit, also referred to as a geosynchronous equatorial orbit[ a ] (GEO), is a circular geosynchronous orbit 35,786 km (22,236 mi) in altitude above Earth's equator, 42,164 km (26,199 mi) in radius from Earth's center, and following the direction of Earth's rotation.
Sep 21, 2012 · Thank you for your help.In summary, the radius of the satellite's orbit can be found using the formula d = sqrt((GMm)/F), where G is the gravitational constant, M is the mass of the Earth, m is the mass of the satellite, and F is the force experienced by the satellite.
Let us use the subscript 1 for the Moon and the subscript 2 for the satellite. We are asked to find T 2 T 2. The given information tells us that the orbital radius of the Moon is r 1 = 3. 84 × 10 8 m r 1 = 3. 84 × 10 8 m, and that the period of the Moon is T 1 = 27.3 d T 1 = 27.3 d.
Since the ISS orbits 4.00 x 10 2 km above Earth’s surface, the radius at which it orbits is R E + 4.00 x 10 2 km. We use Equations \ref{13.7} and \ref{13.8} to find the orbital speed and period, respectively.
