Question

A spacecraft in the shape of a long cylinder has a length of 100 m, and its mass with occupants is 1 480 kg. It has strayed too close to a black hole having a mass 90 times that of the Sun. The nose of the spacecraft points toward the black hole, and the distance between the nose and the center of the black hole is 10.0 km.
What is the difference in the gravitational field acting on the occupants in the nose of the ship and on those in the rear of the ship, farthest from the black hole? This difference in acceleration grows rapidly as the ship approaches the black hole. It puts the body of the ship under extreme tension and eventually tears it apart.

Answer 1

Answer:

2352645198509.9604 m/s²

Explanation:

G = Gravitational constant = 6.67 × 10⁻¹¹ m³/kgs²

M = Mass of black hole = $90\times 1.989\times 10^{30}\ kg$

$R_{f}$ = 10000+100 m

$R_{sb}$ = Distance between the nose and the center of the black hole = 10000 m

The difference in the gravitational field in this system is given by

$\Delta F=\dfrac{GMm}{R_{f}^2}-\dfrac{GMm}{R_{sb}^2}\\\Rightarrow \Delta g=GM\left(\frac{1}{R_f^2}-\frac{1}{R_{sb}^2}\right)\\\Rightarrow g=6.67\times 10^{-11}\times 90\times 1.989\times 10^{30}\left(\frac{1}{(10000+100)^2}-\frac{1}{10000^2}\right)\\\Rightarrow \Delta g=-2352645198509.9604\ m/s^2$

The acceleration is 2352645198509.9604 m/s²