A satellite is in a circular orbit about the earth (ME = 5.98 1024 kg). The period of the satellite is 3.70 104 s. What is the speed at which the satellite travels?
T=2•π•R•sqrt(R/G•M)
the gravitational constant G =6.67•10^-11 N•m²/kg²,
M= 5.985•10^24 kg
T²=4π²R³/G•M
Solve for R.
V=2πR/T
To find the speed at which the satellite travels, we can use the formula for the speed of an object in circular orbit. The formula is:
v = (2πr) / T
Where:
v = speed of the satellite
π = Pi (approximately 3.14159)
r = radius of the orbit
T = period of the satellite
In this case, we already know the period of the satellite is 3.70 * 10^4 seconds.
To find the radius of the orbit, we can use the formula for the circumference of a circle:
C = 2πr
Where:
C = circumference of the orbit
π = Pi
r = radius of the orbit
The formula for the circumference of a circle can be rearranged to solve for the radius:
r = C / (2π)
The circumference of the orbit is equal to the circumference of the circular path followed by the satellite. Since it is a circular orbit around the Earth, the circumference of the orbit is the circumference of the Earth's equator, which is approximately 40,075 km or 4.008 * 10^7 m.
Let's calculate the speed of the satellite step by step:
Step 1: Calculate the radius of the orbit
r = C / (2π) = 4.008 * 10^7 m / (2 * 3.14159) ~= 6.39 * 10^6 m
Step 2: Calculate the speed of the satellite using the formula
v = (2πr) / T = (2 * 3.14159 * 6.39 * 10^6 m) / (3.70 * 10^4 s) ~= 1.73 * 10^3 m/s
Therefore, the speed at which the satellite travels is approximately 1.73 * 10^3 m/s.