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1.	Calculate the book value of gold in the iron meteorite Lutetia (Science Brief 13.7.1) based on the following information: Diameter = 115 km, density = 7.90 g/cm , concentration of gold = 3 0.6 µg/g (ppm), price of gold = $12,405 per kg (Kargel, 1994).
Answer» Volume of Lutetita (V): V = (4 x π x r³)/3 = (4 x 3.14 x (57.5)³)/3 = 0.7959 x 10⁶km³ V = 0.7959 × 10⁶ × 10¹⁵ = 0.7959 × 10²¹cm³ Mass of Lutetia (M): M = Volume × Density M = 0. 7959 × 10²¹ 7.90 = 6.2876 × 10²¹g Mass of gold in Lutetia (Au): Au = Mass × Concentration Au = 6.2876 × 10²¹ × 0.6 = 3.77 × 10²¹µg Au = (3.77 x 10²¹)/10⁶ = 3.77 x 10¹⁵g = 3.77 x 10¹²kg Value of the gold ($): $ = 3.77 × 10¹² × 12,405 = 4.67 × 10¹⁶ Value of the gold = $4.67 × 10¹⁶ US

Calculate the book value of gold in the iron meteorite Lutetia (Science Brief 13.7.1) based on the following information: Diameter = 115 km, density = 7.90 g/cm , concentration of gold = 3 0.6 µg/g (ppm), price of gold = $12,405 per kg (Kargel, 1994).

Answer»

Volume of Lutetita (V):

V = (4 x π x r³)/3 = (4 x 3.14 x (57.5)³)/3 = 0.7959 x 10⁶km³

V = 0.7959 × 10⁶ × 10¹⁵ = 0.7959 × 10²¹cm³

Mass of Lutetia (M):

M = Volume × Density

M = 0. 7959 × 10²¹ 7.90 = 6.2876 × 10²¹g

Mass of gold in Lutetia (Au):

Au = Mass × Concentration

Au = 6.2876 × 10²¹ × 0.6 = 3.77 × 10²¹µg

Au = (3.77 x 10²¹)/10⁶ = 3.77 x 10¹⁵g = 3.77 x 10¹²kg

Value of the gold ($):

$ = 3.77 × 10¹² × 12,405 = 4.67 × 10¹⁶

Value of the gold = $4.67 × 10¹⁶ US

Discussion

2.	The orbit of the Apollo asteroid Geographos has an average radius a = 1.24 AU and an eccentricity e = 0.34. Calculate the distance of this asteroid from the Sun at perihelion (q) and at aphelion (Q).
Answer» e = f/a, f = e x a f = 0.34 × 1.24 = 0.4216 AU q = a - f = 1.24 - 0.42 = 0.82 AU Q = a + f = 1.24 + 0.42 = 1.66 AU

The orbit of the Apollo asteroid Geographos has an average radius a = 1.24 AU and an eccentricity e = 0.34. Calculate the distance of this asteroid from the Sun at perihelion (q) and at aphelion (Q).

Answer»

e = f/a, f = e x a

f = 0.34 × 1.24 = 0.4216 AU

q = a - f = 1.24 - 0.42 = 0.82 AU

Q = a + f = 1.24 + 0.42 = 1.66 AU

Discussion

3.	Estimate the average distance between asteroids that reside between the orbits of Mars and Jupiter. Total area = 1.74 x 1018 km2. Assume that this area is populated by 2 × 106 individual asteroids and that the area surrounding each asteroid is circular in shape.
Answer» Area surrounding each asteroid (A) A((1.74 x 10¹⁸)/(2 x 10⁶)) = 0.87 x 10¹²km² Radius of the area (A) A = r²π, r = (A/π)^1/2 r = ((0.87 x 10¹²)/3.14)^1/2 = (0.2770 x 10¹²)^1/2 = 0.526 x 10⁶km r = 526, 000km

Estimate the average distance between asteroids that reside between the orbits of Mars and Jupiter. Total area = 1.74 x 1018 km2. Assume that this area is populated by 2 × 106 individual asteroids and that the area surrounding each asteroid is circular in shape.

Answer»

Area surrounding each asteroid (A)

A((1.74 x 10¹⁸)/(2 x 10⁶)) = 0.87 x 10¹²km²

Radius of the area (A)

A = r²π, r = (A/π)^1/2

r = ((0.87 x 10¹²)/3.14)^1/2 = (0.2770 x 10¹²)^1/2 = 0.526 x 10⁶km

r = 526, 000km

Discussion