After the Sun exhausts its nuclear fuel, its ultimate fate may be to collapse to a white dwarf state. In this state, it would have approximately the same mass as it has now, but its radius would be equal to the radius of the Earth.
(a) Calculate the average density of the white dwarf.
kg/m3
(b) Calculate the surface free-fall acceleration.
m/s2
(c) Calculate the gravitational potential energy associated with a 4.62-kg object at the surface of the white dwarf.
J
To calculate the average density of the white dwarf, you need to understand its mass and volume. You already know that the white dwarf would have the same mass as the Sun, and its radius would be equal to the radius of the Earth. Since the mass is the same, we can use the mass of the Sun, which is approximately 1.989 × 10^30 kilograms.
(a) Average Density:
To calculate the average density, you need to divide the mass by the volume. The volume of a sphere can be calculated using the formula:
V = (4/3)πr^3
Rearranging the formula to solve for the radius:
r = (3V/(4π))^(1/3)
Substituting the given radius of the Earth:
r = (3(4/3)π(6.371 × 10^6)^3 /(4π))^(1/3)
Simplifying the equation:
r ≈ 6.371 × 10^6 meters
Now, you can calculate the average density using the mass and volume:
Density = mass / volume
Density ≈ (1.989 × 10^30 kg) / ((4/3)π(6.371 × 10^6)^3 m^3)
Simplifying the equation:
Density ≈ 1.989 × 10^30 / (4/3 × 3.14159 × 6.371^3) kg/m^3
Calculating the value:
Density ≈ 5.6 × 10^6 kg/m^3
Therefore, the average density of the white dwarf is approximately 5.6 × 10^6 kg/m^3.
(b) Surface Free-fall Acceleration:
The surface free-fall acceleration can be calculated using the formula:
g = G * M / R^2
Where:
G is the gravitational constant (approximately 6.674 × 10^-11 N(m/kg)^2)
M is the mass of the white dwarf (same as the Sun's mass, approximately 1.989 × 10^30 kg)
R is the radius of the white dwarf (same as the Earth's radius, approximately 6.371 × 10^6 m)
Substituting the given values into the formula:
g = 6.674 × 10^-11 * 1.989 × 10^30 / (6.371 × 10^6)^2
Calculating the value:
g ≈ 3.6 × 10^11 m/s^2
Therefore, the surface free-fall acceleration of the white dwarf is approximately 3.6 × 10^11 m/s^2.
(c) Gravitational Potential Energy:
The gravitational potential energy can be calculated using the formula:
PE = - G * M * m / R
Where:
G is the gravitational constant (approximately 6.674 × 10^-11 N(m/kg)^2)
M is the mass of the white dwarf (same as the Sun's mass, approximately 1.989 × 10^30 kg)
m is the object's mass (4.62 kg)
R is the radius of the white dwarf (same as the Earth's radius, approximately 6.371 × 10^6 m)
Substituting the given values into the formula:
PE = - 6.674 × 10^-11 * 1.989 × 10^30 * 4.62 / 6.371 × 10^6
Calculating the value:
PE ≈ - 5.61 × 10^12 J
Therefore, the gravitational potential energy associated with a 4.62 kg object at the surface of the white dwarf is approximately -5.61 × 10^12 Joules. Note that the negative sign indicates that the potential energy is negative, which implies a bound state.