Question

What is the velocity of an Earth satellite that is in a circular orbit with a radius of 7.5 × 107 meters measured from the center of the Earth?

Answer 1

Is there a formula?

Answer 2

Centripetal force is balanced by the gravitational force
\[{\cancel{m}v^2\over (R_e+h)}=G\frac{M_e\cancel{m}}{(R_e+h)^2}\]
m = mass of satellite (note that velocity becomes independent of this mass)
Me = mass of earth
Re = Radius of earth
h = heoght above Earth's surface,
r = Re+h = \(7.5\times10^7\)m
G = gravitational const.

Answer 3

how do you do this?!! D; @aum

Answer 4

We can assume the weight of the satellite is approximately the same on the surface of the Earth as it is in the orbit.

The centripetal force on the satellite is balanced by the weight of the satellite:
\(\Large \cancel {m}g = \huge \frac{\cancel{m}v^2}{r}\)
\(\large v^2 = rg; ~~~~v = \sqrt{rg} = \sqrt{7.5 \times 10^7 * 9.8} \Large \frac{m}{s} = ?\)

Answer 5

are you sure???.....i got a weird numer /.\

Answer 6

268384053.2

Answer 7

:/ i dont understand......

Answer 8

\[
\sqrt{7.5 \times 10^7 * 9.8} \frac{m}{s} = \sqrt{735000000} = 27,111 \frac{m}{s}
\]

Answer 9

soo..2.9x10^3??

Answer 10

No. Approximately, \(2.7 \times 10^4\) m/s

Answer 11

What are the answer choices?

Answer 12

2.9x10^3
2.3x10^3
7.5x10^3
5.9x10^3

Answer 13

Then they want us to use the more accurate formula provided by the previous person:
\[
{\cancel{m}v^2\over (R_e+h)}=G\frac{M_e\cancel{m}}{(R_e+h)^2} \\
v^2 = G\frac{M_e}{(R_e+h)} \\
v = \sqrt{G\frac{M_e}{(R_e+h)}} \\
G = 6.673 \times 10^{-11} \frac{Nm^2}{Kg^2}; ~~~~
M_e = 5.98 \times 10^{24}~Kg; ~~~~
R_e + h = 7.5 \times 10^7 ~m \\
v = \sqrt{G\frac{M_e}{(R_e+h)}} =
\sqrt{6.673 \times 10^{-11}\frac{5.98 \times 10^{24}}{(7.5\times10^7)}} =
\sqrt{5.3206 \times 10^{6}} = 2.3 \times 10^3~m/s
\]

Answer 14

omg!! your soo smart!!! xD thank you! i got the answer correct :)

Answer 15

You are welcome.

Answer 16

:)