# Assume that a satellite orbits Earth 249 km above its surface. Given that the mass of earth is 5.97 x 10^24 kgAnd the radius of Earth is 6.38 x 10^6m, what is the satellite's speed?

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### 1 Answer

To solve for the satellite's speed, apply the formula below which is based on Netwon's Law.

`v= sqrt((GM)/R) `

where,

v - velocity of satellite moving about a central body in circular motion,

G - gravitational constant, `6.673 xx 10^(-11)` `(N*m^2)/(kg^2)`

M - mass of the central body, and

R - distance from the center of M to the satellite

Since, the satellite is 249 km `(0.249xx10^6m)` above the earth's surface and the radius of the earth is `6.38xx10^6 m` , the value of R is

`R= (0.249xx10^6)+(6.38xx10^6)= 6.629xx10^6`

Then, substitute `G=6.673 xx 10^(-11) (N*m^2)/(kg^2)` , `M=5.97xx10^24 kg` and `R=6.629xx10^6m` to the formula.

`v=sqrt (((6.673 xx 10^(-11))*(5.97xx10^24))/(6.629xx10^6))`

`v=7752.18`

**Hence, the velocity of the satellite is 7752.18 m/s.**