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:

link

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.

explanation:

could i give you a exaplo that could you?

the box does not move because you need to use a force greater than 300 n. you are only using 50 n. it needs to at least be 301 n for the box to move because then it is an unbalanced force and it will move the way that the greater force is pushing it. if you use only 300 n it will be a balanced force and will not move because the force you are exerting on the   box is the same as its force.

hope this !

im sorry if i went too in depth, i am going to include a link to a video on khan academy that will explain this.   hope that you like the video and that it explains more in detail to you.

that is the link.