Question

for a satellite to orbit earth at a constant distance, its centripetal acceleration must be equal to earths gravitational acceleration. if a satellite is to orbit with a constant circular radius of 8,000,000 m, what is the approximate required velocity of the satellite? (recall that earth has a mass of 5.97×10^24 kg and g = 6.67×10^(-11) n·m^2/kg^2.) a. 8023 m/s b. 10,628 m/s c. 7055 m/s d. 9840 m/s

Explanation:

Step1: Equate centripetal and gravitational acceleration formulas

The centripetal acceleration $a_c=\frac{v^{2}}{r}$, and the gravitational acceleration $a_g = \frac{GM}{r^{2}}$. Since $a_c=a_g$, we have $\frac{v^{2}}{r}=\frac{GM}{r^{2}}$.

Step2: Solve for velocity $v$

From $\frac{v^{2}}{r}=\frac{GM}{r^{2}}$, we can cross - multiply to get $v^{2}=\frac{GM}{r}$, then $v=\sqrt{\frac{GM}{r}}$.
Substitute $G = 6.67\times10^{-11}\ N\cdot m^{2}/kg^{2}$, $M = 5.97\times 10^{24}\ kg$, and $r=8000000\ m$ into the formula.
$v=\sqrt{\frac{6.67\times 10^{-11}\times5.97\times 10^{24}}{8000000}}$
$v=\sqrt{\frac{6.67\times5.97\times 10^{-11 + 24}}{8\times10^{6}}}$
$v=\sqrt{\frac{39.8299\times 10^{13}}{8\times10^{6}}}$
$v=\sqrt{4.9787375\times 10^{7}}$
$v\approx7055\ m/s$

Answer:

C. 7055 m/s