A satellite that is 4175 miles from the center of the earth, orbits with a period of 90 minutes. What is its centripetal acceleration?
If I recall correctly,
a = rω²
where ω = angular velocity
Here, ω = 2π/1.5 hr
a = 4175 * (2π/1.5)² = 4175 * 17.546 = 73254 mi/hr²
Look good to you?
To find the centripetal acceleration of the satellite, we can use the formula:
a = 4π²r / T²
where:
- a is the centripetal acceleration,
- r is the distance of the satellite from the center of the earth,
- T is the period of the satellite's orbit.
Given that the distance of the satellite from the center of the earth is 4175 miles and the period of the orbit is 90 minutes, we can substitute these values into the formula to calculate the centripetal acceleration.
Plugging in the values:
a = 4π²(4175) / (90)²
Calculating this:
a ≈ 5.31 miles per hour per hour
Therefore, the centripetal acceleration of the satellite is approximately 5.31 miles per hour per hour.
To find the centripetal acceleration of a satellite, we can use the formula:
ac = (4π²r) / T²
where ac is the centripetal acceleration, r is the radius of the orbit, and T is the period of the orbit.
In this case, the radius of the orbit, r, is given as 4175 miles, and the period of the orbit, T, is given as 90 minutes. However, we need to convert the period to seconds since the formula requires the time to be in seconds.
1 minute = 60 seconds
Therefore, the period, T, in seconds is:
T = 90 minutes * 60 seconds/minute = 5400 seconds
Now, we can calculate the centripetal acceleration using the formula:
ac = (4π² * 4175 miles) / (5400 seconds)²
To calculate the value, let's assume π is equal to 3.14:
ac = (4 * 3.14² * 4175 miles) / (5400 seconds)²
Now, we can simplify and calculate the value:
ac ≈ 0.087 miles/second²
Therefore, the centripetal acceleration of the satellite is approximately 0.087 miles/second².