Let's break down and solve each part: --- ### **(a) Show the tidal force formula** The tidal force on a small object of mass \( m \) is defined as the difference in the gravitational force at the near side and far side of the object. The gravitational force at a distance \( r \) from mass \( M \): \[ F_g = \frac{GMm}{r^2} \] Let the distance from the center of the massive object to the near side be \( R - \Delta r/2 \) and to the far side be \( R + \Delta r/2 \), where \( \Delta r \) is the length of the object. The force at the near side: \[ F_{\text{near}} = \frac{GMm}{(R - \Delta r/2)^2} \] The force at the far side: \[ F_{\text{far}} = \frac{GMm}{(R + \Delta r/2)^2} \] The tidal force is the difference: \[ F_{\text{tidal}} = F_{\text{near}} - F_{\text{far}} \] For \( \Delta r \ll R \), use a Taylor expansion: \[ f(R \pm \Delta r/2) \approx f(R) \pm \frac{\Delta r}{2} f'(R) \] So, \[ F_{\text{tidal}} \approx \frac{d}{dr}\left(\frac{GMm}{r^2}\right) \Bigg|_{r=R} \cdot \Delta r \] \[ \frac{d}{dr}\left(\frac{GMm}{r^2}\right) = -2\frac{GMm}{r^3} \] So, \[ F_{\text{tidal}} \approx 2\frac{GMm}{R^3} \Delta r \] --- ### **(b) Tidal force for the supermassive black hole** Given: - Mass of BH, \( M = 4 \times 10^6 \) solar masses - Solar mass, \( M_\odot = 1.99 \times 10^{30} \) kg - Your mass, \( m = 70 \) kg - Your height, \( \Delta r = 2.0 \) m First, calculate Schwarzschild radius: \[ R_S = \frac{2GM}{c^2} \] where \( c = 3.0 \times 10^8 \) m/s. Total mass: \[ M = 4 \times 10^6 \times 1.99 \times 10^{30} = 7.96 \times 10^{36} \text{ kg} \] \[ R_S = \frac{2 \times 6.67 \times 10^{-11} \times 7.96 \times 10^{36}}{(3.0 \times 10^8)^2} \] \[ = \frac{1.061 \times 10^{27}}{9 \times 10^{16}} \] \[ = 1.18 \times 10^{10} \text{ m} \] Now, calculate the tidal force at \( R_S \): \[ F_{\text{tidal}} = 2 \frac{GMm}{R^3} \Delta r \] Set \( R = R_S \): \[ F_{\text{tidal}} = 2 \frac{6.67 \times 10^{-11} \times 7.96 \times 10^{36} \times 70}{(1.18 \times 10^{10})^3} \times 2 \] Calculate numerator: \[ 6.67 \times 10^{-11} \times 7.96 \times 10^{36} \times 70 \times 2 = 6.67 \times 7.96 \times 70 \times 2 \times 10^{25} \] \[ = 6.67 \times 7.96 = 53.08 \] \[ 53.08 \times 70 = 3715.6 \] \[ 3715.6 \times 2 = 7431.2 \] \[ \rightarrow 7.4312 \times 10^{3} \times 10^{25} = 7.43 \times 10^{28} \] Denominator: \[ (1.18 \times 10^{10})^3 = 1.64 \times 10^{30} \] So, \[ F_{\text{tidal}} = \frac{7.43 \times 10^{28}}{1.64 \times 10^{30}} \approx 0.045 \text{ N} \] **This is a very small tidal force!** At the event horizon of a supermassive black hole, you would not be "spaghettified"—you would survive crossing the event horizon. --- ### **(c) Tidal force for a 10-solar-mass black hole** Now, - \( M = 10 \times 1.99 \times 10^{30} = 1.99 \times 10^{31} \) kg Schwarzschild radius: \[ R_S = \frac{2 \times 6.67 \times 10^{-11} \times 1.99 \times 10^{31}}{9 \times 10^{16}} \] \[ = \frac{2.653 \times 10^{21}}{9 \times 10^{16}} = 2.95 \times 10^{4} \text{ m} \] Now calculate tidal force: \[ F_{\text{tidal}} = 2 \frac{GMm}{R^3} \Delta r \] \[ = 2 \times \frac{6.67 \times 10^{-11} \times 1.99 \times 10^{31} \times 70}{(2.95 \times 10^4)^3} \times 2 \] Numerator: \[ 6.67 \times 10^{-11} \times 1.99 \times 10^{31} \times 70 \times 2 = 6.67 \times 1.99 = 13.28 \] \[ 13.28 \times 70 = 929.6 \] \[ 929.6 \times 2 = 1859.2 \] \[ = 1.86 \times 10^3 \times 10^{20} = 1.86 \times 10^{23} \] Denominator: \[ (2.95 \times 10^4)^3 = 2.57 \times 10^{14} \] So, \[ F_{\text{tidal}} = \frac{1.86 \times 10^{23}}{2.57 \times 10^{14}} = 7.24 \times 10^8 \text{ N} \] **This is an enormous tidal force!** You would be torn apart ("spaghettified") well before reaching the event horizon of the smaller black hole. --- ## **Summary Table** | Black Hole | Tidal Force at Event Horizon | Survive? | |---------------------|-----------------------------|--------------------| | Supermassive (center of galaxy) | ~0.045 N | **Yes** | | 10-solar-mass | \( 7.24 \times 10^8 \) N | **No** (spaghettified) | --- Let me know if you want to see the full step-by-step algebra for the derivatives in part (a)!