# A satelite located in it orbit 3200km from the center of the earth, moves twice round the earth after 48 hours. What is the uniform velocity?

### 7 Answers

- billrussell42Lv 71 year agoFavorite Answer
twice IN 48 hours? ie, the period is 24 hours?

24 hr = 86400 x

circumference of orbit = 2π3.2e6 m = 2.011e7 m

2.011e7 m / 86400 s = 233 m/s

edit: looking again, that satellite is inside the earth!! Your numbers are wrong.

edit, I suspect you mean 3200 km above the surface. Then R = 3200 km + 6,371 km = 9.57e6 meters

plug that into the equations above....

- Andrew SmithLv 71 year ago
ZERO.

Velocity is the change of displacement / change in time.

If it has completed two complete revolutions around a stationary earth then the velocity ( relative to the earth ) is zero because there is no displacement.

It has a displacement relative to the sun equivalent to the motion of the earth around the sun.

And displacements relative to the galaxy etc.

Yet the presumption is that you wanted the VELOCITY RELATIVE TO THE EARTH.

You cannot use velocity as a synonym for SPEED. If the questioner had asked for speed the answer would have been different.

PS applying Kepler's laws I think that your questioner meant 32 THOUSAND km from the centre of the earth. In my head I got that 27 thousand km gives an orbit of 12 hours.

Hence to get an orbit of 24 hours would be further out.

You wrote 3.2 thousand or one tenth of what I think was the intended value.

- jeffrcalLv 71 year ago
The earth's circumference is 40,000 km (The kilometer was originally defined as one ten-thousandths of the distance from the pole to the equator on a line drawn through Paris.). If we divide that by 2*pi to get the radius of the earth that is: 6,369 km.

In short, therefore, something that is 3200 km from the center of the earth is not a satellite in orbit around the earth rather it is buried over 3000 km under the surface.

- NCSLv 71 year ago
Assume 3200 km from the SURFACE of the Earth...

For "orbit", centripetal acceleration = gravitational acceleration, or

ω²r = (2π/T)²r = 4π²r / T² = v²/r = GM/r²

where the first four terms are all expressions for centripetal acceleration

and G = Newton's gravitational constant = 6.674e−11 N·m²/kg²

and M = mass of central body

and r = orbit radius

and T = period

and v = orbit velocity

so

(2π/T)²r = v²/r

(2π/T)²r² = v²

2πr / T = v = 2π * (3200+6400)e3m / (24*3600s) = 698 m/s

NOT 466 m/s

Your data is inconsistent. In order to have a period of 24 hours ("twice... after 48 hours") the radius would have to be such that

(2π/T)²r = GM/r²

r³ = GMT² / 4π² = 6.674e−11N·m²/kg² * 5.98e24kg * (24*3600s)² / 4π² = 7.55e22 m²

r = 4.2e7 m

or about 42000 km from the Earth's center, or about 36000 km above the surface of the Earth. Not 3200 km from the center.

If the satellite was 3200 km above the surface of the Earth, the period would be such that

(2π/T)²r = GM/r²

T² = 2πr³ / GM = 2π * (6.371e6m + 3.2e6m)³ / (6.674e−11N·m²/kg² * 5.98e24kg)

T² = 1.4e7 s²

T = 3715 s

or about 1 h 2 minutes, or almost 48 times after 48 hours.

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- Anonymous1 year ago
over 9000 mph!

- az_lenderLv 71 year ago
A "satellite" that is 3200 km from the center of the earth is about 3000 km underground !