For a satellite to orbit Earth at a constant distance, its centrifugal acceleration must be equal and opposite Earth's gravitational acceleration. If a satellite is to orbit at a constant distance from Earth at a circular radius of \( 8,300,000 \mathrm{~m} \), what is the required velocity of the satellite? (Assume the acceleration due to Earth's gravity is \( 5.6 \mathrm{~m} / \mathrm{s}^{2} \) at this altitude.)

To find the required velocity of a satellite orbiting Earth at a constant distance, we can use the relationship between gravitational acceleration and centripetal acceleration. The gravitational acceleration \( g \) at the altitude of the satellite is given as \( 5.6 \, \text{m/s}^2 \). For a satellite in circular orbit, the centripetal acceleration \( a_c \) is provided by the gravitational force and is given by the formula: \[ a_c = \frac{v^2}{r} \] where: - \( v \) is the orbital velocity of the satellite, - \( r \) is the radius of the orbit (the distance from the center of the Earth to the satellite). Since the satellite is in a stable orbit, the centripetal acceleration must equal the gravitational acceleration: \[ \frac{v^2}{r} = g \] Rearranging this equation to solve for \( v \): \[ v^2 = g \cdot r \] \[ v = \sqrt{g \cdot r} \] Now, substituting the values for \( g \) and \( r \): - \( g = 5.6 \, \text{m/s}^2 \) - \( r = 8,300,000 \, \text{m} \) Calculating \( v \): \[ v = \sqrt{5.6 \, \text{m/s}^2 \cdot 8,300,000 \, \text{m}} \] Calculating the product: \[ 5.6 \cdot 8,300,000 = 46,480,000 \] Now taking the square root: \[ v = \sqrt{46,480,000} \approx 6815.5 \, \text{m/s} \] Thus, the required velocity of the satellite is approximately: \[ \boxed{6815.5 \, \text{m/s}} \]