GPS satellites are at altitudes of exactly 20000 km, where the acceleartion of gravity is only 5.8% of its surface value.What is the orbital speed of the GPS satelites to the nearest hour.
I got 9774, but I know this is wrong any help would be greatly appreciated.
Early recognition of a geostationary orbit was made by the Russian Konstantin Tsiolkovsky early in the 20th century. Others referred to the unique orbit in writings about space travel, space stations, and communications. It was probably Arthur C. Clarke who was given the major credit for the use of this orbit for the purpose of worldwide communications.
The geostationary orbit is one where a spacecraft or satellite appears to hover over a fixed point on the Earth's surface. There is only one geostationary orbit in contrast to there being many geosynchronous orbits. What is the difference you ask? Well, a geosycnchronous orbit is one with a period equal to the earth's rotational period, which, contrary to popular belief, is 23hr-56min-4.09sec., not 24 hours. Thus, the required altltude providing this period is ~22,238 miles, or ~35,788 kilometers. An orbit with this period and altitude can exist at any inclination to the equator but clearly, a satellite in any such orbit with an inclination to the equator, cannot remain over a fixed point on the Earth's surface. On the other hand, a satellite in an orbit in the plane of the earth's equator and with the required altitude and period, does remain fixed over a point on the equator. This equatorial geosynchronous orbit is what is referred to as a geostationary orbit. The orbital velocity of satellites in this orbit is ~10,088 feet per second or ~6,877 MPH. The point on the orbit where the circular velocity of the launching rocket reaches 10,088 fps, and shuts down, is the point where the separated satellite will remain. This very popular orbit is saturated with TV broadcast, government, international, maritime, meteorological, military, domestic and shortwave radio satellites. At the current time, ~270 satellites are on active status in this orbit.
To find the orbital speed of the GPS satellites, we can use the formula for orbital velocity:
v = sqrt(G * M / r)
where v is the orbital velocity, G is the gravitational constant, M is the mass of Earth, and r is the radius or distance from the satellite to the center of the Earth.
In this case, the altitude of the satellites is given as 20,000 km, but we need to convert it to meters for consistency. 1 km is equal to 1000 meters, so 20,000 km is 20,000 * 1000 = 20,000,000 meters.
Given that the acceleration of gravity at this altitude is only 5.8% of its surface value, we can calculate the effective gravitational force as follows:
g_effective = g_surface * (5.8 / 100)
= 9.8 m/s^2 * (5.8 / 100)
= 0.568 m/s^2
Now, we can plug in the values into the formula:
v = sqrt(G * M / r)
v = sqrt(6.67 * 10^-11 N m^2/kg^2 * (5.97 * 10^24 kg) / (20,000,000 m))
v = sqrt(1.19 * 10^12 N m^2/kg^2)
v ≈ 1.09 * 10^6 m/s
Since the question asks for the answer in the nearest hour, let's convert this to hours:
v_hour = v / (3600 s/hour)
v_hour = (1.09 * 10^6 m/s) / (3600 s/hour)
v_hour ≈ 303.06 m/hour
Rounding this to the nearest hour, the orbital speed of GPS satellites is approximately 303 m/hour.