Final answer:
The orbital speed of a spaceship just above the surface of a white dwarf is approximately 1.15 x 10^5 m/s. The orbital period would be approximately 1 minute.
Step-by-step explanation:
In order to find the orbital speed and orbital period of a spaceship in orbit just above the surface of a white dwarf, we need to apply the principles of gravitational force and centripetal force. The formula for orbital speed is given by:
v = sqrt(G * m / r)
where v is the orbital speed, G is the gravitational constant (~6.674 × 10^-11 N m^2 / kg^2), m is the mass of the white dwarf, and r is the distance between the spaceship and the center of the white dwarf.
The formula for orbital period is given by:
T = 2 * pi * r / v
where T is the orbital period. Now, let's calculate the orbital speed first.
From the given information, we know that the white dwarf has the mass of the Sun compressed into the size of the Earth. So, the mass of the white dwarf would be equal to the mass of the Sun, which is approximately 2 x 10^30 kg. The radius of the white dwarf is given as 1.4% of the Sun's radius, which is about 3.5 x 10^6 m. Plugging these values into the formula for orbital speed, we get:
v = sqrt((6.674 × 10^-11 N m^2 / kg^2) * (2 x 10^30 kg) / (3.5 x 10^6 m))
Calculating this gives us a value of v = 1.15 x 10^5 m/s.
Next, let's calculate the orbital period using the value of v we just found.
Plugging the values into the formula for the orbital period, we get:
T = 2 * pi * (3.5 x 10^6 m) / (1.15 x 10^5 m/s)
Calculating this gives us a value of T = 60.9 s or approximately 1 minute.