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A spacecraft in the shape of a long cylinder has a length of 100 m, and its mass with occupants is 1,310 kg. It has strayed too close to a black hole with a mass 105 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.

Please calculate any gravitational effects or potential forces acting on the spacecraft due to its proximity to the black hole.

Answer :

(a) The total force on the spacecraft is 1.80 x 10^17 N

(b)The difference in the gravitational fields acting on the occupants in the nose of the ship and on those in the rear of the ship, farthest from the black hole is 3.97 x 10^12 m/s^2

Forces are influences that can change the movement of an object. A force can change the velocity of an object with mass, i.e. accelerate it. Forces can also be described intuitively by pushing or pulling. A force has both magnitude and direction and is a vector quantity.

solution:

(a) To get the total force acting on the craft, we have to divide it into slices of thickness "dr" so that the mass of one slice is dm = (mass of craft / length of craft) dr = 10.6 dr

force on one slice = G M dm / r^2 where M is mass of black hole

total force on craft = integral of ( G M dm / r^2) = GM * integral of (10.6 dr/r^2) from r = 10000 m to r = 10100 m.

total force = 10.6 GM (1/10000 - 1/10100) =

= 10.6 * 6.67x10^-11 * 105 * 1.99x10^30 * (1/10000 - 1/10100) =

= 1.80 x 10^17 Newtons

(b) strength of field at front of craft =

= G M / r^2 = 6.67x10^-11 * 105 * 1.99x10^30 / 10000^2 =

= 1.3936 x 10^14 m/s^2

at back of craft =

= G M / r^2 = 6.67x10^-11 * 105 * 1.99x10^30 / 10100^2 =

= 1.3662 x 10^14 m/s^2

difference? 1.3936 - 1.3662 = 3.97x 10^12 m/s^2

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