A satellite of mass 500kg orbits the Earth in a circular path with radius 1.0e^7 m .
1) What is the period of the satellite?
I know that the equation is
T= 2(pi)r^3/2 / sqrt GmE
mE = 5.98e^24
does G = 6.67e^-11 ??
T=2•π•R•sqrt(R/Gm)
the gravitational constant G =6.67•10^-11 N•m²/kg²,
Earth’s mass is M = 5.97•10^24 kg,
Earth’s radius is R = 6.378•10^6 m
How did you get 6.378e^6 for R?
From reference book
Yes, you are correct. The equation for the period of a satellite in a circular orbit around a celestial body is:
T = (2π√r^3) / √(GmE)
Where:
- T is the period of the satellite (the time it takes for one complete orbit)
- r is the radius of the orbit
- G is the gravitational constant (approximately equal to 6.67 × 10^-11 N m^2/kg^2)
- mE is the mass of the Earth (approximately equal to 5.98 × 10^24 kg)
So in this case, you can substitute the given values into the equation to find the period of the satellite:
T = (2π√(1.0 × 10^7)^3) / √(6.67 × 10^-11 × 5.98 × 10^24)
Now you just need to plug in the values and calculate the result.
T = (2π√(1.0 × 10^7)^3) / √(6.67 × 10^-11 × 5.98 × 10^24)
T = (2π√1.0 × 10^21) / √(6.67 × 5.98 × 10^13)
T = (2π1.0 × 10^10) / √(6.67 × 5.98 × 10^13)
T = (2 × 3.14 × 10^10) / √(6.67 × 5.98 × 10^13)
Now you can calculate the value:
T ≈ 6.28 × 10^10 / √(3.99126 × 10^27)
T ≈ 6.28 × 10^10 / 6.3195 × 10^13
T ≈ 9.94 × 10^-4 seconds
Therefore, the period of the satellite is approximately 9.94 × 10^-4 seconds.