Communications satellites are placed in a circular orbit where they stay directly over a fixed point on the equator as the earth rotates. These are called geosynchronous orbits. The altitude of a geosynchronous orbit is 3.58×107m(˜22,000miles).
Part A
What is the period of a satellite in a geosynchronous orbit?
Express your answer to three significant figures and include the appropriate units.
Part B
Find the value of g at this altitude.
Express your answer to three significant figures and include the appropriate units.
Part C
What is the weight of a 2000 kg satellite in a geosynchronous orbit?
Express your answer as an integer and include the appropriate units.
The correct answer and explanation is :
Part A: Period of a Geosynchronous Satellite
To find the period ( T ) of a satellite in a geosynchronous orbit, we recognize that such a satellite must orbit the Earth once every 24 hours, to remain over the same spot on the equator.
But we convert this to seconds:
[
T = 24 \, \text{hours} \times 3600 \, \text{seconds/hour} = \boxed{8.64 \times 10^4 \, \text{s}}
]
Part B: Gravitational Acceleration ( g ) at the Altitude
We use the formula for gravitational acceleration at a distance ( r ) from the center of the Earth:
[
g = \frac{GM}{r^2}
]
Where:
- ( G = 6.674 \times 10^{-11} \, \text{m}^3/\text{kg} \cdot \text{s}^2 )
- ( M = 5.972 \times 10^{24} \, \text{kg} )
- ( r = R_{\text{Earth}} + h = 6.371 \times 10^6 \, \text{m} + 3.58 \times 10^7 \, \text{m} = 4.217 \times 10^7 \, \text{m} )
Now plug in:
[
g = \frac{6.674 \times 10^{-11} \cdot 5.972 \times 10^{24}}{(4.217 \times 10^7)^2}
= \boxed{0.224 \, \text{m/s}^2}
]
Part C: Weight of a 2000 kg Satellite
Weight ( W ) is given by:
[
W = mg = 2000 \, \text{kg} \cdot 0.224 \, \text{m/s}^2 = \boxed{448 \, \text{N}}
]
Explanation (300 words)
Geosynchronous satellites are essential for consistent communication because they appear stationary from the Earth’s surface. To achieve this, they must orbit at the same rotational rate as Earth—once every 24 hours. This gives the satellite a period ( T ) of 86,400 seconds.
The altitude for such an orbit is about 35,800 km above Earth’s surface. However, to calculate gravitational effects like ( g ), we need the distance from the Earth’s center, which is the sum of Earth’s radius and the satellite’s altitude. The gravitational force decreases with distance from Earth’s center, so ( g ) is smaller at this height compared to Earth’s surface. Using Newton’s Law of Universal Gravitation, we calculate ( g ) at that distance to be approximately 0.224 m/s²—far less than the 9.81 m/s² we experience on the surface.
Weight is the gravitational force acting on an object. So, for a satellite with a mass of 2000 kg, its weight in orbit is the product of mass and the weaker gravitational field: ( 2000 \times 0.224 = 448 \, \text{N} ). While this is much less than its Earth-surface weight (which would be 19,620 N), it still experiences some gravitational pull, which is why it remains in orbit rather than flying off into space.
Even though the satellite is “weightless” in the sense that it is in free fall, it’s still under the influence of gravity. This is key to orbital motion—it’s the continuous “fall” around the Earth that maintains its circular path.
