A satellite orbiting Earth has a tangential velocity of 5000 m/s. Earth's mass is 6 * 1024 kg and its radius is 6.4 * 106 m. The distance of the satellite from Earth, written in standard notation, is * blank * m.

1 x 10^7 meters The equation for the orbital velocity of a satellite is v = sqrt (u (2/r-1/a)) where v = velocity r = radius at satellite's location a = semi-major axis of satellite's orbit u = standard gravitational parameter (product of G and M) G = 6.67408 Ă - 10-11 m^3 / (kg s^2) so u = 6.67408 Ă - 10^-11 m^3 / (kg s^2) * 6 Ă - 10^24 kg = 4.004448 x 10^14 m^3/s^2 Since the orbit in this problem is assumed to be circular, we can simplify v = sqrt (u (2/r-1/a)) to v = sqrt (u/r) because r and a have the same value. So plugging in our known values and solving for r, we get v = sqrt (u/r) 5000 m/s = sqrt (4.004448 x 10^14 m^3/s^2 / r) 25000000 m^2/s^2 = 4.004448 x 10^14 m^3/s^2 / r 25000000 m^2/s^2 r = 4.004448 x 10^14 m^3/s^2 r = 4.004448 x 10^14 m^3/s^2 / 25000000 m^2/s^2 r = 4.004448 x 10^14 / 25000000 m r = 16017792 m The distance calculated above is from the center of the earth, so to get the altitude above earth, we need to subtract the earth's radius. So a = 16017792 m - 6.4 x 10^6 m = 9.62 x 10^6 Since our input data only has 1 significant figure, the result rounded to 1 significant digit is 1 x 10^7