Scientists want to place a 3100 kg satellite in orbit around Mars. They plan to have the satellite orbit a distance equal to 1.9 times the radius of Mars above the surface of the planet. Here is some information that will help solve this problem:
mmars = 6.4191 x 1023 kg
rmars = 3.397 x 106 m
G = 6.67428 x 10-11 N-m2/kg2
1)What should the radius of the orbit be (measured from the center of Mars), if we want the satellite to take 8 times longer to complete one full revolution of its orbit?
so far I have found speed to orbit 2085m/s and 8.2hrs to complete one revolution
To solve this problem, we can use the concept of centripetal force and gravitational force to find the radius of the orbit.
1) First, let's find the speed at which the satellite should orbit. We can use the concept of centripetal force:
Centripetal force = Gravitational force
The centripetal force acting on the satellite is provided by the gravitational force between Mars and the satellite:
mv^2/r = (G*mmars*msatellite)/r^2
Where:
m = mass of the satellite (3100 kg)
v = velocity of the satellite
r = radius of the orbit (distance from the center of Mars)
G = gravitational constant (6.67428 x 10^-11 N-m^2/kg^2)
mmars = mass of Mars (6.4191 x 10^23 kg)
2) Given that the satellite should take 8 times longer to complete one full revolution, we can relate the time it takes to complete one revolution to the orbital period:
Time for one revolution = 8 * Orbital period
The orbital period (T) is the time it takes for one complete revolution of the satellite around Mars.
T = (2*pi*r) / v
Where:
r = radius of the orbit (distance from the center of Mars)
v = velocity of the satellite
Now we have two equations, one involving the forces and one involving the time. We can solve these equations to find the radius of the orbit (r).
First, let's calculate the speed of the satellite:
mv^2/r = (G*mmars*msatellite)/r^2
Rearranging the equation, we can solve for v:
v = sqrt((G*mmars*msatellite)/r)
Substituting the given values:
v = sqrt((6.67428 x 10^-11 N-m^2/kg^2 * 6.4191 x 10^23 kg * 3100 kg) / r)
Now, we can use this value for v in the equation for the orbital period:
T = (2*pi*r) / sqrt((6.67428 x 10^-11 N-m^2/kg^2 * 6.4191 x 10^23 kg * 3100 kg) / r)
Since we want the satellite to take 8 times longer to complete one full revolution, we can substitute 8T for the time in the equation:
8T = (2*pi*r) / sqrt((6.67428 x 10^-11 N-m^2/kg^2 * 6.4191 x 10^23 kg * 3100 kg) / r)
Simplifying the equation:
8T = (2*pi*r) / sqrt((6.67428 x 10^-11 N-m^2/kg^2 * 6.4191 x 10^23 kg * 3100 kg) / r)
Multiply both sides of the equation by sqrt((6.67428 x 10^-11 N-m^2/kg^2 * 6.4191 x 10^23 kg * 3100 kg) / r):
8T * sqrt((6.67428 x 10^-11 N-m^2/kg^2 * 6.4191 x 10^23 kg * 3100 kg) / r) = 2*pi*r
Squaring both sides:
(8T * sqrt((6.67428 x 10^-11 N-m^2/kg^2 * 6.4191 x 10^23 kg * 3100 kg) / r))^2 = (2*pi*r)^2
Now we have an equation with only one unknown, r. We can rearrange and solve for r:
r = (2*pi)^2 * r^3 / (8T * sqrt((6.67428 x 10^-11 N-m^2/kg^2 * 6.4191 x 10^23 kg * 3100 kg))
Now, substitute the given values for T and solve for r:
r = (2*pi)^2 * r^3 / (8 * (8.2 * 3600) * sqrt((6.67428 x 10^-11 N-m^2/kg^2 * 6.4191 x 10^23 kg * 3100 kg))
object is 2.9 times radius from center. Remember to add Mars radius to altitude. I do not think you need the mass of the object.
F/m = a = = v^2/R = G Mmars / R^2 = omega^2 R