Geosynchronous satellites orbit the Earth at an altitude of about 3.58×10^7 m. Given that the Earth's radius is 6.38×10^6 m and its mass is 5.97×10^24 kg, what is the magnitude of the gravitational acceleration at the altitude of one of these satellites?
Please explain as well as give an answer. I would like to know the hows and whys for future reference.
g=force/mass=GMe/distance^2 is one way, if you look up the mass of the Earth
Another way is to compute centripetal acceleration
a=v^2/r = w^2/r where w=2PI/period and r is given above.
and period is a day in seconds.
To find the magnitude of the gravitational acceleration at the altitude of a geosynchronous satellite, we need to use the formula for gravitational acceleration. The formula is given by:
g = G * (M / r^2)
Where:
g is the gravitational acceleration,
G is the universal gravitational constant (6.67430 x 10^-11 N(m/kg)^2),
M is the mass of the Earth,
and r is the distance between the center of the Earth and the satellite.
In this case, we are given the altitude of the satellite (3.58 x 10^7 m) and the radius of the Earth (6.38 x 10^6 m). We can calculate the distance between the center of the Earth and the satellite (r) as follows:
r = altitude + radius
r = (3.58 x 10^7 m) + (6.38 x 10^6 m)
r = 4.22 x 10^7 m
Now, we substitute the values into the gravitational acceleration formula:
g = (6.67430 x 10^-11 N(m/kg)^2) * (5.97 x 10^24 kg) / (4.22 x 10^7 m)^2
Simplifying the equation:
g = (39.8 N) / (1.78 x 10^15 m^2)
g = 2.24 x 10^-5 m/s^2
Therefore, the magnitude of the gravitational acceleration at the altitude of a geosynchronous satellite is approximately 2.24 x 10^-5 m/s^2.
It is worth noting that this value is significantly smaller than the gravitational acceleration at the Earth's surface (9.8 m/s^2) because the force of gravity decreases with distance from the center of the Earth.