What is the average density and free-fall acceleration of a white dwarf?

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Homework Statement



After a star like the Sun has exhausted most of the hydrogen in its core it expands and cools to form a red giant. Eventually, when it has exhausted all its nuclear fuel, it sheds its outer layers and contracts and becomes a white dwarf of similar size to the Earth as shown below. Note that the mass of the sun is 2 × 1030 kg, the radius of the Earth is 6,380 km and Newton's gravitational constant G is 6.67 × 10–11 N m2 kg–2.

Consider a white dwarf of 0.650 solar mass and 0.500 Earth radii.

(a) Calculate the average density of the white dwarf

(b) Calculate the free-fall acceleration on the surface of the white dwarf

The Attempt at a Solution



(a) I believe the average density is given by mass/volume

mass is [tex]0.650 \times (2 \times 10^{30})=1.3 \times 10^{30}[/tex]

volume is [tex]\frac{4}{3}\pi (0.5 \times 6380)^3 = 1.35 \times 10^{11}[/tex]

m/v=9.6 × 1018

But the correct answer is 9560000000 kg/m³. I appreciate it if anyone could show me what's wrong with my working :confused:
 
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You used the Earth radius in km, so you calculated the density in kg/km^3, not kg/m^3.