A satellite is orbiting the earth just above its surface. The centripetal force making the satellite follow a circular trajectory is just its weight, so its centripetal acceleration is about 9.81 m/s2 (the acceleration due to gravity near the earth's surface). If the earth's radius is about 6360 km, how fast must the satellite be moving? How long will it take for the satellite to complete one trip around the earth?
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To determine the speed of the satellite, we need to use the centripetal force equation:
F = (mv^2)/r
where F is the gravitational force acting as the centripetal force, m is the mass of the satellite, v is the velocity of the satellite, and r is the radius of the Earth.
In this case, we can substitute the weight of the satellite for the centripetal force, as they are equal:
Weight = mg = (mv^2)/r
Rearranging the equation gives:
v^2 = g * r
Substituting the given values:
v^2 = 9.81 m/s^2 * 6360 km
v^2 = 9.81 m/s^2 * 6360000 m
v^2 = 62,373,600 m^2/s^2
Taking the square root of both sides:
v ≈ √(62,373,600 m^2/s^2) ≈ 7900 m/s
So, the satellite must be moving at approximately 7900 m/s.
To calculate the time it takes for the satellite to complete one orbit around the Earth, we can use the formula:
T = (2 * π * r) / v
where T is the period, r is the radius of the Earth, and v is the velocity of the satellite.
Substituting the given values:
T = (2 * π * 6360 km) / 7900 m/s
T = (2 * π * 6360000 m) / 7900 m/s
Calculating, we have:
T ≈ 5071 s ≈ 1.41 hours
Therefore, it will take approximately 5071 seconds or 1.41 hours for the satellite to complete one orbit around the Earth.
To find the speed at which the satellite must be moving, we can use the formula for centripetal acceleration:
a = v^2 / r
Where:
- a is the centripetal acceleration
- v is the velocity of the satellite
- r is the radius of the circular path (earth's radius plus height of the satellite)
Given that the centripetal acceleration is 9.81 m/s^2 and the radius of the earth is 6360 km (which we need to convert to meters), we can rearrange the formula to solve for v:
v = √(a * r)
Let's plug in the values:
r = 6360 km = 6360 * 1000 m = 6,360,000 m
a = 9.81 m/s^2
v = √(9.81 * 6,360,000)
Now let's calculate the value of v:
v ≈ 7,910 m/s
Therefore, the satellite must be moving at approximately 7,910 m/s to maintain its orbit.
To calculate the time taken for the satellite to complete one trip around the earth (orbit period), we can use the formula:
T = (2π * r) / v
Where:
- T is the time taken
- r is the radius of the circular path (earth's radius plus height of the satellite)
- v is the velocity of the satellite
Using the values we have:
r = 6,360,000 m
v = 7,910 m/s
T = (2π * 6,360,000) / 7,910
Now let's calculate the value of T:
T ≈ 5066 seconds
Therefore, it will take approximately 5066 seconds for the satellite to complete one trip around the earth.