Question

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.

Answer 1

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

Answer 2

Answer:

$9.6 \cdot 10^6 m$

Explanation:

we can solve the problem by equalizing the force of gravity that keeps the satellite in circular motion with the centripetal force:

$G\frac{m M}{r^2}=m\frac{v^2}{r}$

where

$G=6.67 \cdot 10^{-11}$ is the gravitational constant

m is the satellite's mass

M is the Earth's mass

r is the distance of the satellite from Earth's center

v is the tangential velocity of the satellite

By re-arranging the formula and substituting the numbers, we find r:

$r=G \frac{m}{v^2}=(6.67 \cdot 10^{-11})\frac{6 \cdot 10^{24}}{(5000)^2}=1.6 \cdot 10^7 m$

However, this is the distance of the satellite from Earth's center. Since we want to know its distance from the Earth's surface, we must subtract the value of the Earth's radius:

$r'=r-R=1.6 \cdot 10^7 m-6.4 \cdot 10^6 m=9.6 \cdot 10^6 m$