Radio signals travel at a rate of 3 × 10^8 meters per second. How many seconds would it take for a radio signal to travel from a satellite to the surface of Earth if the satellite is orbiting at a height of 3.6 × 10^7 meters? (Hint: Time is distance divided by rate.)
(1 point)
Responses
8.3 seconds
8.3 seconds
1.2 × 10^–1 seconds
1.2 × 10 ^–1 seconds
1.08 × 10^16 seconds
1.08 × 10 ^16 seconds
10.8 × 10^15 seconds
The distance the radio signal has to travel is the height of the satellite plus the radius of the Earth. The radius of the Earth is approximately 6.37 x 10^6 meters.
Total distance = height of satellite + radius of Earth
Total distance = 3.6 x 10^7 m + 6.37 x 10^6 m
Total distance = 3.6 x 10^7 m + 6.37 x 10^6 m
Total distance = 4.237 x 10^7 m
Time = distance / rate
Time = 4.237 x 10^7 m / 3 x 10^8 m/s
Time = 1.41233 seconds
Therefore, it would take 1.41 seconds for a radio signal to travel from a satellite to the surface of Earth.
please choose one of the options
1.2 × 10^–1 seconds
are you sure
My previous response was incorrect. Let me recalculate:
Total distance = height of satellite + radius of Earth
Total distance = 3.6 x 10^7 m + 6.37 x 10^6 m
Total distance = 3.6 x 10^7 m + 6.37 x 10^6 m
Total distance = 4.237 x 10^7 m
Time = distance / rate
Time = 4.237 x 10^7 m / 3 x 10^8 m/s
Time = 0.141233 seconds
Therefore, it would take 0.141233 seconds for a radio signal to travel from a satellite to the surface of Earth.
The closest option is 1.2 × 10^–1 seconds, which is equivalent to 0.12 seconds. Thank you for your patience.