A satellite is on orbit 35600 km above the surface of the earth. its angular velocity is 7.25×10–5 rad/sec. What is the vrlocity of the satellite?(The radius of the earth is 6400 km

The velocity of a satellite with an orbit 35600 km above the surface of the earth and an angular velocity of 7.25×10–5 rad/sec = 3045 m/s

Velocity: This is the rate of change of displacement.

To solve this problem we need to use the formula for calculating the velocity of  an object in circular motion

The Velocity of the satellite is given as

V = ωr Equation 1

where V = velocity of the satellite, ω = angular velocity of the satellite, r = radius of the circle.

Note: r = height of the satellite above the surface of the earth+Radius of the earth

From the question,

Given: ω = 7.25×10⁻⁵ rad/sec, r = 35600+6400 = 42000 km = 4.2×10⁷ m

Substitute these values into equation 1

V = (7.25×10⁻⁵)(4.2×10⁷ )

V = 30.45×10²

V = 3045 m/s

Hence, the velocity of the satellite is 3045 m/s

Learn more about velocity here: link

3.045 km/s

Explanation:

When an object is doing a circular motion, the velocity of the object is written as;

v = r*w

where;

r = radius of the circle

w = angular velocity.

In this case, we know that:

w = 7.25*10^(-5) s^-1

And the radius will be equal to the radius of the Earth, plus the height of the satellite, this is:

R = radius of the Earth + 35600 km = 6400km + 35600 km  = 42000 km

Then the velocity of the satellite will be:

v =  42000 km*7.25*10^(-5) s^-1 = 3.045 km/s

snell's law

refractve index = sin incidence / sin refraction