Question

QUESTION 17 The International Space Station which presently has mass of about 420,000 kilograms, maintains an orbit with an altitude of between 330 and 435 kilometers (205 and 270 miles respectively) above the Earth surface by means of re-boost maneuvers using the engines of the Zvezda module or visiting spacecraft to compensate for 2 km/month orbital decay due to the atmospheric drag. Calculate the following for the low orbit of the ISS (330 kilometers above the Earth surface), assuming that it is in a circular orbit the magnitude of the gravitational force exerted by the Earth on the International Space Station: A. Newtons (use scientific notation and round the coefficient to two decimals) the magnitude of the acceleration due to Earth's gravity at the International Space Station B. m/s (round your answer to two decimals) - orbital velocity of the International Space Station C. km/s (round your answer to two decimals) the orbital period of the International Space Station D. minutes (round your answer to the nearest integer) the escape velocity from Earth's gravity for a spacecraft launched from the orbit of the International Space Station E km/s (round your answer to two decimals) Earth's mean radius is RE = 6.37 *106 m; Earth's mass is ME = 5.97x1024kg.

Answer 1

Assuming the circular orbit of 330 Km above the earth surface,

(i) The magnitude of gravitational force exerted by earth on the ISS is

$F_{grav}=\frac{GMm}{\left (R+h \right )^{2}}=\frac{6.67\times 10^{-11}\times 5.97\times 10^{24}\times 420000}{\left ( 6371+330 \right )^{2}\times 10^6}=3.7\times 10^6 \ N$

(ii) Acceleration due to earth gravity at ISS is

$g_{ISS}=\frac{GM}{\left (R+h \right )^{2}}=\frac{6.67\times 10^{-11}\times 5.97\times 10^{24}}{\left ( 6371+330 \right )^{2}\times 10^6}=8.8679\ m/s^{2}$

(iii) The orbital velocity of ISS will be

$v_{ISS}=\sqrt{\frac{GM}{\left (R+h \right )}}=\sqrt{\frac{6.67\times 10^{-11}\times 5.97\times 10^{24}}{\left ( 6371+330 \right )\times 10^3}}=7708.68\ m/s=7.709 \ Km/s$

(iv) The orbital period from Kepler´s law will be

$T_{ISS}=2\pi \sqrt{\frac{\left (R+h \right )^{3}}{GM}}=2\pi \sqrt{\frac{\left ( 6371+330 \right )^{3}\times 10^9}{6.67\times 10^{-11}\times 5.97\times 10^{24}}}=5461.8 \ s=91.031\ min$

(v) Escape velocity from ISS will be

$v_{e,ISS}=\sqrt{\frac{2GM}{R+h}}=\sqrt{2}v_{ISS}=\sqrt{2}\times7.709=10.9\ Km/s$