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3 years ago

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An interstellar space probe is launched from Earth. After a brief period of acceleration it moves with a constant velocity, 70.0

% of the speed of light. Its nuclear-powered batteries supply the energy to keep its data transmitter active continuously. The batteries have a lifetime of 15.9 years as measured in a rest frame. (a) How long do the batteries on the space probe last as measured by mission control on Earth?
yr
(b) How far is the probe from Earth when its batteries fail, as measured by mission control?
ly
(c) How far is the probe from Earth, as measured by its built-in trip odometer; when its batteries fail?
ly

1 answer:

sleet_krkn [62]3 years ago

3 0

Answer:

22.26 years

, 15.585 light years , 11.13 light years

Explanation:

a)

$t' = t/(\sqrt{1-(v/(c*v)/c)}$

= $15.9/\sqrt{(1-0.7*0.7)}$

= 22.26 years

b)

0.7*c*22.26 years

=15.585 light years

c)

0.7*c*15.9

=11.13 light years

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=> $\Delta K = \frac{1}{2} * m * \frac{1}{4} * v_o ^2 - \frac{1}{2} * m * v_o^2$

And $\Delta P$ is the change in electric potential energy from initial position to a position of distance R from the nucleus , this is mathematically represented as

$\Delta P = P_f - P_i$

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$\Delta P = k * \frac{q_1 * q_2 }{R} - 0$

So

$\frac{1}{2} * m * \frac{1}{4} * v_o ^2 - \frac{1}{2} * m * v_o^2 = k * \frac{q_1 * q_2 }{R} - 0$

=> $\frac{1}{2} * m *v_0^2 [ \frac{1}{4} -1 ] = k * \frac{q_1 * q_2 }{R}$

=> $- \frac{3}{8} * m *v_0^2 = k * \frac{q_1 * q_2 }{R} ---(1 )$

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$\Delta K_f = \Delta P_f$

Here $\Delta K$ is the change in kinetic energy from initial position to a position of distance R from the nucleus , this is mathematically represented as

$\Delta K_f = K_f - K_i$

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=> $\Delta K_f = \frac{1}{2} * m * \frac{1}{16} * v_o ^2 - \frac{1}{2} * m * v_o^2$

And $\Delta P$ is the change in electric potential energy from initial position to a position of distance $R_f$ from the nucleus , this is mathematically represented as

$\Delta P_f = P_f - P_i$

Here $P_i$ is zero because the electric potential energy at the initial stage is zero so

$\Delta P_f = k * \frac{q_1 * q_2 }{R_f } - 0$

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=> $\frac{1}{2} * m *v_o^2 [-\frac{15}{16} ] = k * \frac{q_1 * q_2 }{R_f }$

=> $- \frac{15}{32} * m *v_o^2 = k * \frac{q_1 * q_2 }{R_f } ---(2)$

Divide equation 2 by equation 1

$\frac{- \frac{15}{32} * m *v_o^2 }{- \frac{3}{8} * m *v_0^2 } } = \frac{k * \frac{q_1 * q_2 }{R_f } }{k * \frac{q_1 * q_2 }{R } }}$

=> $-\frac{15}{32 } * -\frac{8}{3} = \frac{R}{R_f}$

=> $\frac{5}{4} = \frac{R}{R_f}$

=> $R_f = \frac{4}{5} R$

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