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?
wrong plus ratio
radius * (gravity * 1000) = x
square root(x) then divide by 1000 = velocity
time = 2*pi* 6360 / velocity / 60 for minutes
Your welcome
To find the velocity of the satellite, we can use the formula for centripetal acceleration:
a = v^2 / r
where:
a = centripetal acceleration (9.81 m/s^2)
v = velocity of the satellite (unknown)
r = radius of the earth (6360 km or 6,360,000 meters)
Rearranging the formula, we have:
v^2 = a * r
Plugging in the given values, we have:
v^2 = 9.81 m/s^2 * 6,360,000 meters
Simplifying, we get:
v^2 = 62,301,600 m^2/s^2
To solve for v, we take the square root of both sides:
v = √62,301,600 m^2/s^2
v ≈ 7884.69 m/s
Therefore, the satellite must be moving at approximately 7884.69 m/s.
To find the time it takes for the satellite to complete one trip around the Earth, we can use the formula for the period of a circular motion:
T = 2πr / v
where:
T = period of the satellite (unknown)
r = radius of the earth (6360 km or 6,360,000 meters)
v = velocity of the satellite (7884.69 m/s)
Plugging in the given values, we have:
T = 2π * 6,360,000 meters / 7884.69 m/s
Simplifying, we get:
T ≈ 5072.94 seconds
Therefore, it will take approximately 5072.94 seconds (or about 84.55 minutes) for the satellite to complete one trip around the Earth.
To find the speed at which the satellite must be moving, we can start by calculating the centripetal acceleration using the formula:
a = v^2 / r,
where:
- a is the centripetal acceleration,
- v is the velocity of the satellite,
- r is the radius of the Earth.
In this case, the centripetal acceleration is given as 9.81 m/s^2, and the radius of the Earth is 6360 km or 6,360,000 meters.
Substituting these values into the formula, we have:
9.81 = v^2 / 6,360,000.
We can rearrange the formula to solve for v:
v^2 = 9.81 * 6,360,000.
Taking the square root of both sides, we find:
v = √(9.81 * 6,360,000).
Evaluating this expression, we get:
v ≈ 7905 m/s (rounded to three significant figures).
So, the satellite must be moving at a speed of approximately 7905 m/s.
Next, to determine the time it takes for the satellite to complete one orbit around the Earth (known as the orbital period), we can divide the circumference of the orbit by the satellite's speed:
Time = Circumference / Speed.
The circumference of a circle is given by 2πr, where r is the radius of the orbit. Since the orbit is just above the Earth's surface, we can use the Earth's radius (6,360,000 meters) as the radius of the orbit.
Substituting the values, we have:
Time = (2π * 6,360,000) / 7905.
Evaluating this expression, we find:
Time ≈ 5063 seconds (rounded to the nearest whole number).
Therefore, it would take approximately 5063 seconds for the satellite to complete one trip around the Earth.