Question

Zero, a hypothetical planet, has a mass of 5.7 x 1023 kg, a radius of 3.0 x 106 m, and no atmosphere. A 10 kg space probe is to be launched vertically from its surface. (a) If the probe is launched with an initial kinetic energy of 5.0 x 107 J, what will be its kinetic energy when it is 4.0 x 106 m from the center of Zero? (b) If the probe is to achieve a maximum distance of 8.0 x 106 m from the center of Zero, with what initial kinetic energy must it be launched from the surface of Zero?

Answer 1

Part a)

`KE = 1.83 * 10^7 J`

Part b)

`KE = 7.92 * 10^7 J`

Step-by-step explanation:

Here by energy conservation we can say

initial total energy = final total energy

now we have

`-(Gm_1m_2)/(r_1) + (1)/(2)mv_1^2 = -(Gm_1m_2)/(r_2) + (1)/(2)mv_2^2`

now we have

`-((6.67* 10^(-11))(5.7 * 10^(23))(10))/(3* 10^6) + 5 * 10^7 = -((6.67 * 10^(-11))(5.7 * 10^(23))(10))/(4 * 10^6) + KE`

now we have

`KE = 1.83 * 10^7 J`

Part b)

Now again at maximum height the final kinetic energy will be zero

so we have

`-(Gm_1m_2)/(r_1) + (1)/(2)mv_1^2 = -(Gm_1m_2)/(r_2) + (1)/(2)mv_2^2`

now we have

`-((6.67* 10^(-11))(5.7 * 10^(23))(10))/(3* 10^6) + KE = -((6.67 * 10^(-11))(5.7 * 10^(23))(10))/(8 * 10^6) + 0`

now we have

`KE = 7.92 * 10^7 J`