Question

(part 1 of 3) A satellite of mass 398 kg is launched from a site on the Equator into an orbit at 544 km above Earth’s surface. If the orbit is circular, what is the satellite’s speed in orbit? The gravitational constant is 6.67259 × 10^−11 N · m2/kg2 , the mass of the earth is 5.98 × 10^24 kg and its radius is 6.37 × 10^6 m. Answer in units of m/s.

(part 2 of 3) What is the orbital period of this satellite? Answer in units of h.

(part 3 of 3) What is the minimum energy necessary to place this satellite in orbit, assuming no air friction? Answer in units of J.

Answer 1

Solution)

Part 1) speed,  v = sqrt(GM/r) = sqrt(6.67x10^-11*5.98x10^24/(6.37x10^6+ 544x10^3)) = 7595m/s (Ans)

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Part 2)

We know,

GMm/r^2 = m*v^2/r = m*4π^2*r/T^2

so orbital period, T= sqrt(4π^2*r^3/GM) = sqrt(4π^2*(6.37x10^6+ 544x10^3)^3/(6.67x10^-11*5.98x10^24) = 910s =0.25 hrs (Ans)

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Part 3) Minimum Energy = GMm*(1/R - 1/r) + 1/2*m*v^2

= 6.67x10^-11*5.98x10^24*398*(1/(6.37x10^6) - 1/(6.37x10^6 + 544x10^3)) + 1/2*398*(7595)^2

= 9.89 x10^9J

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Good luck!:)