Question

Q C Ganymede is the largest of Jupiter's moons. Consider a rocket on the surface of Ganymede, at the point farthest from the planet (Fig. P13.41). Model the rocket as a parti-cle. (b) Determine the escape speed for the rocket from the planet-satellite system. The radius of Ganymede is 2.64 ×10⁶m , and its mass is 1.495 × 10²³kg . The distance between Jupiter and Ganymede is 1.071 × 10⁹m, and the mass of Jupiter is 1.90 × 10²⁷kg . Ignore the motion of Jupiter and Ganymede as they revolve about their center of mass.

Answer 1

The answer is $v=15.6 \mathrm{~km} / \mathrm{s}$.

What is kinetic energy?

• A particle or an item that is in motion has a sort of energy called kinetic energy. An item accumulates kinetic energy when work, which involves the transfer of energy, is done on it by exerting a net force.
• Kinetic energy comes in five forms: radiant, thermal, acoustic, electrical, and mechanical.
• The energy of a body in motion, or kinetic energy (KE), is essentially the energy of all moving objects. Along with potential energy, which is the stored energy present in objects at rest, it is one of the two primary types of energy.
• Explain that a moving object's mass and speed are two factors that impact the amount of kinetic energy it will possess.

Determine the escape speed for the rocket from the planet-satellite system:

The potential energy of the rocket due to Ganymede when it is on the surface of the Ganymede is,

$U_1=-\frac{G M_{\mathrm{G}} m}{R_G}$

The potential energy of the rocket due to Jupiter

when it is on the surface of the Ganymede is,

$U_2=-\frac{G M_{\mathrm{J}} m}{R}$

Here, R is a separation between Jupiter and Ganymede.

To escape from the surface of Ganymede potential energy of the rocket due to Jupiter and Ganymede is equal to the kinetic energy of the rocket.

$\begin{aligned}&\frac{1}{2} m v^2+U_1+U_2=0 \\&\frac{1}{2} m v^2=-U_1-U_2 \\&\frac{1}{2} m v^2=\frac{G M_{\mathrm{G}} m}{R_G}+\frac{G M_{\mathrm{J}} m}{R} \\&\frac{1}{2} v^2=\frac{G M_{\mathrm{G}}}{R_G}+\frac{G M_{\mathrm{J}}}{R}\end{aligned}$

$v^2=\frac{2 G M_{\mathrm{G}}}{R_G}+\frac{2 G M_{\mathrm{T}}}{R}$

$v=\sqrt{2 G\left(\frac{M_{\mathrm{G}}}{R_G}+\frac{M}{R}\right)}$

$v=\sqrt{2\left(6.67 \times 10^{-11} \mathrm{~N} \cdot \mathrm{m}^2 / \mathrm{kg}^2\right)\left(\frac{1.495 \times 10^{23} \mathrm{~kg}}{2.64 \times 10^6 \mathrm{~m}}+\frac{1.90 \times 10^{27} \mathrm{~kg}}{1.071 \times 10^9 \mathrm{~m}}\right)}$

$v=15.6 \times 10^3 \mathrm{~m} / \mathrm{s}$

$v=15.6 \mathrm{~km} / \mathrm{s}$

To learn more about kinetic energy, refer to:

brainly.com/question/25959744

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