Wonder Woman and Superman fly to an altitude of 1690 km , carrying between them a chest full of jewels that they intend to put into orbit around Earth. They want to make this tempting treasure inaccessible to their evil enemies who are trying to gain possession of it, yet keep it available for themselves for future use when they retire and settle down. But perhaps the time to retire is now! They accidentally drop the chest, which leaves their weary hands at rest, and discover that they are no longer capable of catching it as it falls into the Pacific Ocean. At what speed does the chest impact the surface of the water? Ignore air resistance, although in the real world it would make a world of difference. The radius and mass of Earth are 6370 km and 5.98×1024 kg , respectively.

Respuesta :

Answer:

5120 m/s

Explanation:

The acceleration due to gravity is:

g = MG / r²

where M is the mass of the earth, G is the universal constant of gravitation, and r is the distance from the earth's center to the object's center.

Here, r = h + R, where h is the height of the chest above the surface and R is the radius of the earth.

g = MG / (h + R)²

Acceleration is the derivative of velocity:

dv/dt = MG / (h + R)²

Using chain rule, we can say:

(dv/dh) (dh/dt) = MG / (h + R)²

(dv/dh) v = MG / (h + R)²

Separate the variables:

v dv = MG / (h + R)² dh

Integrating:

∫₀ᵛ v dv = MG ∫₀ʰ dh / (h + R)²

½ v² |₀ᵛ = -MG / (h + R) |₀ʰ

½ (v² − 0²) = -MG / (h + R) − -MG / (0 + R)

½ v² = -MG / (h + R) + MG / R

½ v² = MGh / (R(h + R))

v² = 2MGh / (R(h + R))

Given:

M = 5.98×10²⁴ kg

R = 6.37×10⁶ m

h = 1.69×10⁶ m

G = 6.67×10⁻¹¹ m³/kg/s²

Plugging in:

v² = 2 (5.98×10²⁴) (6.67×10⁻¹¹) (1.69×10⁶) / ((6.37×10⁶) (1.69×10⁶ + 6.37×10⁶))

v² = 2 (5.98) (6.67) (1.69) / ((6.37) (1.69 + 6.37)) × 10⁷

v ≈ 5120 m/s

Notice that if we had approximated g as a constant 9.8 m/s², we would have gotten an answer of:

v² = v₀² + 2a(x - x₀)

v² = (0 m/s)² + 2 (9.8 m/s²) (1.69×10⁶ m - 0 m)

v 5760 m/s

So we know that our calculated velocity of 5120 m/s is a reasonable answer.

The Newton's second law  and the law of universal gravitation allows to find the result for the speed of the chest when reaching the ocean is:

  •  The velocity is v = 5120 m / s

The law of universal gravitation is stable that the gravitational force between bodies is attractive and is proportional to the mass of the bodies and inversely proportional to the square of the distance.

         

           [tex]F= - G \frac{Mm}{r^2}[/tex]

Where M and m are the mass of the two bodies and r is the distance.

Indicate the height from where the chest falls h= 1690 km = 1,690 10⁶ m, they also give the radius and the mass of the earth.

Newton's second law establishes a relationship between force, mass and the acceleration of bodies.

           F = m a  

          [tex]- G \frac{Mm}{r^2} = m a \\a= - G \frac{M}{r^2}[/tex]

The distance of the body from the center of the planet is

          r = R + h

The acceleration is defined as the variation of velocity with time.

          [tex]a= \frac{dv}{dt} \\ \frac{dv}{dt} = - G \frac{M}{(h+R)^2}[/tex]

       

Let's use the chain rule

         [tex]\frac{dv}{dh}\ \frac{dh}{dt} = - GM \frac{1}{(h+R)^2 }[/tex]  

The velocity is the derivative of the position with respect to the time.

         [tex]v=\frac{dh}{dt} \\ v \ \frac{dv}{dh} = - GM \ \frac{1}{(h+R)^2}[/tex]    

         

To solve we use the method of separation of variables and we integrate.

       [tex]\int\limits^v_0 {v} \, dv = -GM \int\limits^0_h {\frac{1}{(h+R)^2} } \, dh \\\frac{1}{2} ( v^2 - 0) = -GM (-1) [ \frac{1}{(0+R)} - \frac{1}{(h+R)}] \\\frac{1}{2} v^2 = GM \ \frac{h}{(h+R)R}[/tex]

      [tex]v^2 = 2GM \ \frac{h}{(h+R) R }[/tex]  

Let's calculate

      v² = [tex]2 \ 6.67 \ 10^{-11} \ 5.98 \ 10^{24}} \ \frac{1.690 }{(1.609 + 6.370) 6.370} \ 10^{-6}[/tex]

      v = 5120 m/s

In conclusion we use Newton's second law and the universal gravitation's law we can find the result for the speed of the chest when reaching the ocean is;

  • The velocity is v = 5120 m / s

Learn more about the law of universal gravitation here: brainly.com/question/2347945