You are designing a spacecraft intended to monitor a human expedition to mars (mass 6.42×1023kg, radius 3.39×106m). this spacecraft will orbit around the martian equator with an orbital period of 24.66 h, the same as the rotation period of mars, so that it will always be above the same point on the equator. what must be the radius of the orbit?

 h = 17 10⁶ m from surface of mars

Explanation:

For this exercise we will use Newton's second law where force is the force of universal gravitation

         F = m a

The acceleration is centripetal

         a = v² / r

         G m M / r² = m v² / r

The speed module is constant, so we use the uniform motion ratio

        v = d / t

Where the distance is the length of the circumference and the time is the period of the orbit

         d = 2π r

         v = 2π r / T

We replace

          G M / r² = (4π² r² / T) / r

           r³ = G M T² / 4π²

Let's reduce time to SI units

          T = 24.66 h (3600 s / 1 h) = 88776 s

Let's calculate

         r = ∛ 6.67 10⁻¹¹ 6.42 10²³ 88776² / 4π²

         r = ∛ 8.5485 10²¹ m

         r = 2,045 10⁶ m

This is the distance from the center of the planet, The height, which is the distance from the surface is

        r = + h

        h = r -

        h = 20.45 10 6 - 3.39 106

        h = 17 10⁶ m

see the picture for the vector

the magnitude of the vector by pythagorean theorem

r = √( (-10)^2 + (3)^2 ) = √109

magnitude = √109 units

the direction angle between vector and x-axis is

tan^(-1)(3/10) = 16.699

so the direction is approximately 16.7 degrees north of west

the temperature in kelvin of 2.0 dm3 of an ideal gas is doubled and its pressure is increased by a factor of four. what is the final volume of the gas