Asatellite with a mass of 5.6 e 5 kg is orbiting the earth in a circular path. determine the satellite's velocity if it is orbiting at a distance of 9.8 e 5 m above the earth's surface. earth's mass = 5.98 e 24 kg; earth's radius = 6.357 e 6 m.

This question involves the concepts of centripetal force and gravitational force of attraction.

The velocity of the satellite is "7373.17 m/s".

For the satellite to continue its motion in a circular path its centripetal force must be equal to the gravitational force:

where,

v = speed of satellite = ?

G = universal gravitational constant = 6.67 x 10⁻¹¹ N.m²/kg²

M = Mass of Earth = 5.98 x 10²⁴ kg

r = distance of satellite + radius of earth = 9.8 x 10⁵ m + 6.357 x 10⁶ m

r = 7.337 x 10⁶ m

Therefore,

v = 7373.17 m/s

Learn more about centripetal force here:

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The attached picture shows the centripetal force.

7400 m/s

Explanation:

Centripetal acceleration = gravity

v² / r = GM / r²

v = √(GM / r)

Given:

G = 6.67×10⁻¹¹ m³/kg/s²

M = 5.98×10²⁴ kg

r = 9.8×10⁵ + 6.357×10⁶ = 7.337×10⁶ m

v = √(6.67×10⁻¹¹ (5.98×10²⁴) / (7.337×10⁶))

v = 7400

The orbital velocity is 7400 m/s.

0 = 0.40 / 0 = 0

= 0.20 = 9.80 ⁄2 =?

=0 , = = 0.20 =0.50 0 0.40 ⁄

=0−122 =0−12(9.80 ⁄2)(0.50)2 =1.2