Synchronous communications satellites are placed in a circular orbit that is 3.59 107 m above the surface of the earth. What is the magnitude of the acceleration due to gravity at this distance?
The correct answer and explanation is :
The magnitude of the acceleration due to gravity at a distance of 3.59 × 10⁷ meters above Earth’s surface is approximately 0.223 m/s².
Explanation (300 words):
Satellites in synchronous orbit are positioned far from Earth’s surface, allowing them to match the planet’s rotation period. These satellites orbit at a constant altitude of 3.59 × 10⁷ meters above the Earth. To find the gravitational acceleration at this altitude, we use Newton’s law of universal gravitation:
[
g = \frac{G \cdot M}{r^2}
]
Where:
- ( G ) = 6.674 × 10⁻¹¹ N·m²/kg² (gravitational constant),
- ( M ) = 5.972 × 10²⁴ kg (mass of Earth),
- ( r ) = total distance from the center of Earth = Earth’s radius + satellite altitude,
- ( R ) = 6.371 × 10⁶ m (Earth’s radius),
- ( h ) = 3.59 × 10⁷ m (height above surface),
- Thus, ( r = R + h = 6.371 × 10⁶ + 3.59 × 10⁷ = 4.2271 × 10⁷ ) m.
Plugging in the values:
[
g = \frac{6.674 \times 10^{-11} \cdot 5.972 \times 10^{24}}{(4.2271 \times 10^7)^2}
]
[
g ≈ 0.223 \, \text{m/s}^2
]
This value is much lower than the 9.8 m/s² at Earth’s surface, because gravity decreases with the square of the distance from Earth’s center.
Despite the low gravity, this small force is enough to keep the satellite in orbit, as orbital motion doesn’t require strong gravity—just the right balance between gravitational pull and the satellite’s velocity. Understanding this helps engineers design satellite trajectories and maintain stable geosynchronous communication systems.