A satellite of mass m in a circular orbit at a height of 400Km above the earth's surface. What is the Radius of the Earth (given: 6.35 * 10^6) with respect to the center C of the Earth?

Two masses m1= 35.0 Kg and m2= 15.0Kg are connected by a mass less string over a smooth pulley (no friction), the mass m1 being on a horizontal frictionless table. The motion of the system involves m1 moving horizontally to the right and m2 vertically downward, both with uniform acceleration "a," the string having tension "T."

Using Newton's 2nd law, write down the corresponding equations for the motion of the masses m1 and m2 relating to the tension T and acceleration a and the weight of the mass m2.

What is the relation of the weight of the mass m1 and the reaction force "R" from the surface of the table, assuming that there is a vertical motion of m1.

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You do not appear to have copied the question properly. Why are you asking for the radius of the Earth when they already have told you it is 6.35*10^6 (meters)?

The actual value is 6.378*10^5 m, by the way. Don't forget to include the units (meters), which you failed to do.

If you want the radius of the satellite ORBIT, add 400 km (4*10^5 m) to that.

The actual value is 6.378*10^6 m, by the way.

To find the radius of the Earth with respect to the center, we need to add the altitude of the satellite to the radius of the Earth's surface.

The altitude of the satellite is given as 400 km, and the radius of the Earth's surface is given as 6.35 * 10^6 m.

Given that the satellite is in a circular orbit, the gravitational force acting on it provides the necessary centripetal force to keep it in its orbit. The centripetal force is given by:

Fc = m * (v^2 / r)

where Fc is the centripetal force, m is the mass of the satellite, v is the orbital velocity, and r is the distance from the center of the Earth.

The orbital velocity v can be calculated using the formula:

v = sqrt(G * M / r)

where G is the gravitational constant and M is the mass of the Earth.

Now, equating the centripetal force and gravitational force:

m * (v^2 / r) = G * (m * M) / (r^2)

Cancelling the mass of the satellite (m) on both sides:

(v^2 / r) = G * M / (r^2)

Simplifying further:

v^2 = G * M / r

Substituting the expression for v^2 in the equation for the centripetal force:

m * (G * M / r) = G * (m * M) / (r^2)

Cancelling G and M on both sides:

r^2 = r

Therefore, the radius of the Earth with respect to the center (r) can be found by adding the altitude of the satellite (400 km) to the radius of the Earth's surface (6.35 * 10^6 m).

r = 6.35 * 10^6 m + 400 * 10^3 m

r = 6.75 * 10^6 m

Hence, the radius of the Earth with respect to the center is 6.75 * 10^6 meters.