[amp_mcq option1=”8.04 km\/sec at a height of 285 km” option2=”11.11 km\/sec at a height of 37,400 km” option3=”11.26 km\/sec, the satellite escapes the pull of the earth” option4=”All the above” correct=”option1″]<\/p>\n
The correct answer is: A. 8.04 km\/sec at a height of 285 km.<\/p>\n
A satellite in a circular orbit around the Earth is moving at a speed that is just fast enough to keep it from falling back to Earth. The speed of a satellite in a circular orbit is given by the formula:<\/p>\n
$$v = \\sqrt{\\frac{GM}{r}}$$<\/p>\n
where $M$ is the mass of the Earth, $r$ is the radius of the orbit, and $G$ is the gravitational constant.<\/p>\n
The radius of the Earth is 6371 km, and the radius of a circular orbit at a height of 285 km is 6656 km. Substituting these values into the formula, we get:<\/p>\n
$$v = \\sqrt{\\frac{(6.67408 \\times 10^{-11} \\text{ m}^3 \\text{ kg}^{-1} \\text{ s}^{-2})(5.9722 \\times 10^{24} \\text{ kg})}{6656 \\text{ km}}} = 8.04 \\text{ km\/s}$$<\/p>\n
Therefore, the correct answer is A. 8.04 km\/sec at a height of 285 km.<\/p>\n
Option B is incorrect because the speed of a satellite in a circular orbit at a height of 37,400 km is 11.11 km\/sec.<\/p>\n
Option C is incorrect because the speed of a satellite that escapes the pull of the Earth is 11.26 km\/sec.<\/p>\n","protected":false},"excerpt":{"rendered":"
[amp_mcq option1=”8.04 km\/sec at a height of 285 km” option2=”11.11 km\/sec at a height of 37,400 km” option3=”11.26 km\/sec, the satellite escapes the pull of the earth” option4=”All the above” correct=”option1″]<\/p>\n","protected":false},"author":1,"featured_media":0,"comment_status":"closed","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"footnotes":""},"categories":[645],"tags":[],"class_list":["post-6800","post","type-post","status-publish","format-standard","hentry","category-applied-mechanics-and-graphic-statics","no-featured-image-padding"],"yoast_head":"\n