A 71.4 kg satellite orbits the earth at a distance of 7.00E6 m from the center of the earth (a distance of 630 km above the earth's surface). How fast must the satellite be going to maintain its orbit if the free fall acceleration at this distance is 8.14 m/s^2?
8.14=v^2/r where r= 7e6 m
To find the speed at which the satellite must be moving to maintain its orbit, we can use the concept of centripetal force.
The centripetal force required to keep the satellite in orbit is provided by the gravitational force between the satellite and the Earth. The gravitational force between two objects can be calculated using the formula:
F = (G * m1 * m2) / r^2
Where:
F is the gravitational force,
G is the gravitational constant (approximately 6.67 x 10^-11 N m^2/kg^2),
m1 and m2 are the masses of the two objects, and
r is the distance between the centers of the two objects.
In this case, the satellite is orbiting the Earth, so m1 refers to the mass of the satellite (71.4 kg) and m2 refers to the mass of the Earth (approximately 5.97 x 10^24 kg). The distance between the center of the Earth and the satellite is given as 7.00E6 m.
Plugging in the values into the formula, we get:
F = (6.67 x 10^-11 N m^2/kg^2) * (71.4 kg) * (5.97 x 10^24 kg) / (7.00E6 m)^2
Simplifying the equation, we get:
F = (6.67 x 10^-11 N m^2/kg^2) * (71.4 kg) * (5.97 x 10^24 kg) / (49.0 x 10^12 m^2)
F ≈ 4.02 x 10^16 N
The centripetal force required to maintain the satellite's orbit is equal to the gravitational force between the satellite and the Earth. In other words:
F = m * a
Where:
F is the force,
m is the mass of the satellite, and
a is the centripetal acceleration.
In this case, we know the mass of the satellite is 71.4 kg, and we are given the centripetal acceleration as 8.14 m/s^2.
Plugging in the values, we have:
4.02 x 10^16 N = (71.4 kg) * a
Solving for a gives:
a = (4.02 x 10^16 N) / (71.4 kg)
a ≈ 5.63 x 10^14 m/s^2
The centripetal acceleration is related to the speed of the satellite by the formula:
a = v^2 / r
Where:
a is the centripetal acceleration,
v is the speed of the satellite, and
r is the distance between the satellite and the center of the Earth.
Rearranging the equation to solve for v, we get:
v = sqrt(a * r)
Plugging in the values:
v = sqrt((5.63 x 10^14 m/s^2) * (7.00E6 m))
v ≈ 7.53 x 10^3 m/s
Therefore, the satellite must be moving at a speed of approximately 7.53 x 10^3 m/s to maintain its orbit.