Suppose the Sun could collapse into a neutron star of radius 12.0 km without los

Question

Suppose the Sun could collapse into a neutron star of radius 12.0 km without losing any mass in the process. Your research team is in charge of sending a probe from Earth to study the transformed Sun, and the probe needs to end up in a circular orbit 5400 km from the neutron-Sun's center. (a) Calculate the orbital speed of the probe. b) Later on plans call for construction of a permanent spaceport in that same orbit to study the neutron-Sun in great detail. To transport equipment and supplies, scientists on Earth need you to determine the escape speed for rockets launched from the spaceport (relative to the spaceport) in the direction of the spaceport's orbital velocity at takeoff time. What is that speed? (c) How does it compare to the escape speed at the surface of Earth?

Explanation / Answer

Answers with given data:

(1) Orbital velocity = 4,958,310 m/s, probably report as 5.0E+6 m/s
(2) Station escape velocity = 7,012,110 m/s, probably report as 7.0E+6 m/s
(3) Earth escape velocity = 11,188m/s, so station's velocity is 627x that of Earth

How to find it, using Kepler mechanics

(a) Orbital velocity of a probe over a Sun-mass neutron-star-radius object.

v = SQRT { [GM] / r }

Where

G = Universal Gravitational Constant
M = Mass of Star (Sun mass)
r = Distance from Center

Given

G = 6.67428E-11 m^3/kg-s^2
M = 1.9891E+30 kg
r = 5,400,000 m (SI unit)

Solve

v = SQRT { (6.67428E-11 m^3/kg-s^2) * (1.9891E+30 kg)
v = SQRT { [ 1.328E+20 m^3/s^2 ] / (5,400,000 m) }
v = SQRT { 2.458 m^2/s^2 }
v = 4,958,310 m/s

Perhaps report as 5.0E+6 m/s

(b) Escape velocity from star's orbit at space station

Ve = SQRT { [2GM] / r }
Ve = SQRT { [ 2 * (6.67428E-11 m^3/kg-s^2) * (1.9891E+30 kg) / (5,400,000 m) }
Ve = SQRT { [ 2.655E+20 m^3/s^2 ] / (5,400,000 m) }
Ve = SQRT { 4.970E+13 m^2/s^2 }
Ve = 7,012,110 m/s

Perhaps report as 7.0E+6 m/s

(c) Escape speed from Earth

Same equation, but M = 5.9742E+24 kg, r = 6,371,000 m

Ve = SQRT { [2GM] / r }
Ve = SQRT { [ 2 * (6.67428E-11 m^3/kg-s^2) * (5.9742E+24) ] / (6,371,000 m) }
Ve = SQRT { [ 7.975E+14 m^3/s^2 ] / (6,371,000 m) }
Ve = SQRT { 12,517,1821 m^2/s^2 }
Ve = 11,188m/s

So escape speed from the star-orbiting space station is 627x stronger than Earth

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