When the Sun ends up as a white dwarf, billions of years from now, it will contain about 40 percent of its present-day mass and have the same radius as Earth. What will be its density? How many times more dense is that than the present-day Sun?
We are going to have to have to look up some numbers. You can use Google for that.
Solar mass (now): 1.989*10^30 kg = M1
Solar mass (future): 0.4*M1 = 7.956*10^29 kg = M2
Solar radius (now): 6.96342*10^5 km = R1
Earth radius (now): 6378.1 km
Solar radius (future): 6378.1 km = R2
Solar density increase factor:
(M2/M1)*(R1/R2)^2 = 4769
Future solar density as white dwarf:
= M2/[(4/3)*pi*R3^3]= ____
Your turn.
I used the wrong exponent for R
Solar density increase factor:
(M2/M1)*(R1/R2)^3 = 0.4*1.30*10^6
= 5.2*10^5
Currently, the sun's density is 1.4 times that of water: 1.4 g/cm3.
It will become nearly 1 million times more dense as a white dwarf.
thanks
To calculate the density of the white dwarf, we need to know its mass and volume.
Firstly, we can determine the mass of the white dwarf by multiplying its present-day mass by 0.4 (40 percent). Let's assume the present-day mass of the Sun is 1 solar mass (approximately 2 × 10^30 kg). Therefore, the mass of the white dwarf will be:
1 solar mass * 0.4 = 0.4 solar masses
So, the mass of the white dwarf will be 0.4 times the mass of the present-day Sun.
Next, let's calculate the volume of the white dwarf. Given that its radius will be the same as Earth's radius (approximately 6,371 km), we can use the formula for the volume of a sphere:
Volume = (4/3) * π * (radius)^3
Plugging in the radius:
Volume = (4/3) * π * (6,371 km)^3
The volume will be approximately 1.083 × 10^12 km^3.
Now, we can calculate the density of the white dwarf by dividing its mass by its volume:
Density = mass / volume
Density = (0.4 solar masses * 2 × 10^30 kg) / (1.083 × 10^12 km^3)
Density = 7.4 × 10^11 kg/km^3
So, the density of the white dwarf will be approximately 7.4 × 10^11 kg/km^3.
To determine how many times more dense the white dwarf is compared to the present-day Sun, we need to calculate the density of the Sun. The average density of the Sun is approximately 1.41 × 10^3 kg/km^3.
Therefore, the ratio between the density of the white dwarf and the present-day Sun is:
(7.4 × 10^11 kg/km^3) / (1.41 × 10^3 kg/km^3)
This is approximately 5.25 × 10^8 times more dense than the present-day Sun.
Hence, the density of the white dwarf will be around 7.4 × 10^11 kg/km^3, making it approximately 5.25 × 10^8 times denser than the present-day Sun.