Asatellite orbits the earth a distance of 1.50 × 107 m above the planet's surface and takes 8.65 hours for each revolution about the earth. the earth's radius is 6.38 × 106 m. the acceleration of this satellite is closest

The acceleration of the satellite is

Explanation:

The acceleration in a circular motion is defined as:

 (1)

Where a is the centripetal acceleration, v the velocity and r is the radius.

The equation of the orbital velocity is defined as

(2)

Where r is the radius and T is the period

For this particular case, the radius will be the sum of the high of the satellite () and the Earth radius () :

Then, equation 2 can be used:

⇒

Finally equation 1 can be used:

Hence, the acceleration of the satellite is

The acceleration of the satellite is 0.87 m/s²

The given parameters;

distance of the satellite, r₁ = 1.5 x 10⁷ m

radius of the earth, r₂ = 6.38 x 10⁶ m

time of motion, t = 8.65 hours

The radius of the satellite is calculated as;

R = r₁ + r₂

R = (1.5 x 10⁷) m + (6.38 x 10⁶) m

R = 2.138 x 10⁷ m

The time of motion of the satellite in seconds;

T = 8.65 hours x 3600 s

T = 31,140 s

The velocity of the satellite is  calculated as;

The acceleration of the satellite is  calculated as;

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305.7

explanation:

density =m/v

density =917/3.00

density = 305.

approximately = 305.7

in four significant figures