91Ó°ÊÓ

Given data from example 13.5 and Appendix F can be listed below as,

• Escape velocity is, $V_e=\sqrt{\frac{2GM}{R}}$.
• The mass of Mars is, $M_M=6.42x{10}^{23}\mathrm{kg}$.
• The radius of mars is, $R_M=3.39x{10}^{6}m$.
• The mass of Jupiter is, $M_J=1.90x{10}^{27}\mathrm{kg}$.
• The radius of Jupiter is, $R_J=6.99x{10}^{7}m$

When an item rises in height, its gravitational potential energy increases; it gets smaller as it descends over larger distances.

Because gravity transforms gravitational potential energy to kinetic energy when an item touches the ground, the higher it starts, the quicker it falls.

From example 13.5, escape velocity is given by,

$V_e=\sqrt{\frac{2GM}{R}}$

Here, G is the gravitational constant whose value is $6.67×{10}^{-11}N·m^2/k{g}^2$. M is the mass of the planet, and R is the radius of the planet.

From the above equation, the escape velocity of Mars is given by,

role="math" localid="1668062064598" $V_e=\sqrt{\frac{2G{M}_M}{R_M}}$

Here, G is the gravitational constant, $M_M$is the mass of Mars,$R_M$ is the radius of Mars.

Substitute all values in the above,

$\begin{array}{r}{v}_{eM}=\sqrt{\frac{2\left(6.67×{10}^{−11}\mathrm{N}â‹\dots {\mathrm{m}}^2/{\mathrm{kg}}^2\right)\left(6.42×{10}^{23}\mathrm{kg}\right)}{3.39×{10}^6\mathrm{m}}\left(\frac{1\mathrm{kg}â‹\dots \mathrm{m}/{\mathrm{s}}^2}{1\mathrm{N}}\right)}\\ =\sqrt{\frac{85.64×{10}^{12}{\mathrm{m}}^3/{\mathrm{s}}^2}{3.39×{10}^6\mathrm{m}}}\\ =5.03×{10}^3\mathrm{m}/\mathrm{s}\end{array}$

Thus, the escape velocity of Mars is $5.03×{10}^3\mathrm{m}/\mathrm{s}$.

The escape velocity of Jupiter can be given by,

$V_e=\sqrt{\frac{2G{M}_J}{R_J}}$

Here,G is the gravitational constant,$M_J$ is the mass of Jupiter,$R_J$ is the radius of Jupiter.

Substitute the values in the above equation.

$\begin{array}{r}{v}_{\mathrm{e},}=\sqrt{\frac{2\left(6.67×{10}^{−11}\mathrm{N}â‹\dots {\mathrm{m}}^2/{\mathrm{kg}}^2\right)\left(1.90×{10}^{27}\mathrm{kg}\right)}{6.99×{10}^7\mathrm{m}}\left(\frac{1\mathrm{kg}â‹\dots \mathrm{m}/{\mathrm{s}}^2}{1\mathrm{N}}\right)}\\ =\sqrt{\frac{25.346×{10}^{18}{\mathrm{m}}^3/{\mathrm{s}}^2}{6.99×{10}^7\mathrm{m}}}\\ =6.02×{10}^4\mathrm{m}/\mathrm{s}\end{array}$

Thus, the escape velocity of Jupiter is $6.02×{10}^4\mathrm{m}/\mathrm{s}$.

The kinetic and gravitational potential energies are proportional to the spacecraft's mass. Kinetic and Potential energies can't equal zero unless gravitational forces are present. The mass of the spacecraft will rule out when the kinetic and potential energy become equal leading escape energy independent of the mass of the spacecraft.