a) When a falling meteoroid is at a distance above the Earth\'ssurface of 3.25 t

Question

a) When a falling meteoroid is at a distance above the Earth'ssurface of 3.25 times the Earth's radius, what is its accelerationdue to the Earth's gravitation? ___________ m/s2(towards the earth) incorrect answers i have gotten- 5.39 and .931
b) At the Earth's surface a projectile is launched straight upat a speed of 8.6 km/s. To what height will it rise? Ignore airresistance and the rotation of the Earth. ____________m incorrect answers i have gotten- 3.77 * 10^6 a) When a falling meteoroid is at a distance above the Earth'ssurface of 3.25 times the Earth's radius, what is its accelerationdue to the Earth's gravitation? ___________ m/s2(towards the earth) incorrect answers i have gotten- 5.39 and .931
b) At the Earth's surface a projectile is launched straight upat a speed of 8.6 km/s. To what height will it rise? Ignore airresistance and the rotation of the Earth. ____________m incorrect answers i have gotten- 3.77 * 10^6

Explanation / Answer

height of meteoroid h = 3.25 R where R = radius of earth Accleration due to the earth's gravitation g ' = g * [ R/ ( R + h ) ] ^ 2                                                                   = 0.05536 g                                                                   = 0.05536 * 9.8 m / s^ 2                                                                   = 0.5425 m / s^ 2 (b).Initial speed v = 8.6 km / s= 8600 m / s let the maximum height be H from law of conservation of energy , total energy on surface =total energy at maximum height K.E on surface + P.E on surface = P.E at maximumheight    Since K.E at maximum height is zero     ( 1/ 2) mv ^ 2 + [-GMm / R ] = [-GMm/ ( R + H ) ]        0.5 v^ 2 -[GM /R ] = [-GM/ ( R + H ) ] where G = Gravitational constant = 6.67* 10 ^ -11 N m^2/ kg^ 2            R= Radius = 6.38 * 10 ^ 6 m           M =mass of earth = 5.98 * 10 ^ 24 kg plug the values we get H value

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