Question

Ameteoroidisfirstobservedapproachingtheearthwhenitis402,000kmfromthecenterofthe earth with a true anomaly of 150 . If the speed of the meteoroid at that time is 2.23 km/s, calculate (a) the eccentricity of the trajectory; (b) the altitude at closest approach; and (c) the speed at the closest approach.

Answer 1

Answer:

A. 1.086 B. 5089.8Km C. 8.52Km/s

Explanation:

Energy

$E = \frac{v^{2} }{2} - \frac{u}{(r)} \\\\E = \frac{2.23^{2} }{2} -\frac{398000}{402000} \\\\E = 2.48645 - 0.99\\E = 1.4964 Km^{2}/s^{2}$

$h^{2} = \frac{-1}{2} *\frac{u^{2}* (1 -e^{2} )}{E} \\\\h^{2} = \frac{-1}{2} *\frac{398600^{2}* (1 -e^{2} )}{1.4964}\\\\h^{2} = 5.309*10^{10} * (1 -e^{2} )$

$h^{2} = ur(1 + ecos (theta)) \\\\h^{2} = 398600 * 402000(1 + ecos (150))\\\\h^{2} = 16.02*10^{10} * (1 -0.866e )$

Equating both equations

$5.309*10^{10} * (1 -e^{2} )= 16.02*10^{10} * (1 -0.866e )\\\\1 -e^{2} = 3.018 * (1 -0.866e )\\\\0 = e^{2} + 2.613e - 4.018 \\\\e = -3.697 or 1.0861$

The eccentricity of the trajectory is taking to e the positive value i.e 1.0861

The altitude at closest approach is calculated from any of the two h^2 equations above

$h^{2} = 16.02*10^{10} * (1 -0.866e )\\\\h^{2} = 16.02*10^{10} * (1 -0.866 * 1.086 )\\\\h^{2} = 16.02*10^{10} * 0.059524\\\\h^{2} = 16.02*10^{10} * (1 -0.866e )\\\\h^{2}= 9.5357* 10^{9} Km^{4}/s^{2}$

The perigee radius

$=\frac{h^{2} }{u} *\frac{1}{(1+e )} \\\\=\frac{h^{2} }{u} *\frac{1}{(1+e )} \\\\=\frac{9.5357*10^{9} }{398600} *\frac{1}{(1+1.0861)} \\\\=\frac{h^{2} }{u} *\frac{1}{(1-e^{2} )} \\\\=11467.8Km$

Note: radius of the earth = 6378Km

Perigee altitude = 11467.8 - 6378 =  5089.8 Km

c. speed

$=\frac{h}{Perigee radius} =\frac{\sqrt{9.5357*10^{9} } }{11467.8} \\\\= 8.52 Km/s$