Chapter 7, Problem 13Q

(a)

To determine

The escape speed of a spacecraft from Europa (Jupiter’s moon), which revolves around an orbit of radius $670900\text{ km}$. Given that the spacecraft, after collecting the sample from the satellite’s surface, is returning to Earth.

Answer to Problem 13Q

Solution:

$2.02\text{ km/s}$

Explanation of Solution

Given data:

A spacecraft landed on Jupiter’s moon, which revolves around Jupiter in an orbit of radius $670900\text{ km}$.

Formula used:

The relation for the escape velocity is as:

$v_{escape}={\left(\frac{2GM}{R}\right)}^{\frac{1}{2}}$

Here, $v_{escape}$ is the escape velocity for Europa, G is the universal gravitational constant having value $6.67\times {10}^{-11}\text{ N}\cdot {\text{m}}^{\text{2}}{\text{/kg}}^{\text{2}}$, M is the mass of Europa and R is the radius of Europa.

Explanation:

Mass of Europa is $4.80\times {10}^{22}\text{ kg}$ and its radius is $1.57\times {10}^6\text{ m }$.

Recall the above formula.

$v_{escape}={\left(\frac{2GM}{R}\right)}^{\frac{1}{2}}$

Substitute $4.80\times {10}^{22}\text{ kg}$ for M, $1.57\times {10}^6\text{ m}$ for R and $6.67\times {10}^{-11}\text{ N}\cdot {\text{m}}^{\text{2}}{\text{/kg}}^{\text{2}}$ for G

$\begin{array}{c}{v}_{escape}={\left(\frac{2\left(6.67\times {10}^{-11}{\text{ Nm}}^{\text{2}}{\text{/kg}}^{\text{2}}\right)\left(4.8\times {10}^{22}\text{ kg}\right)}{1.57\times {10}^6\text{ m}}\right)}^{\frac{1}{2}}\\ ={\left(\frac{6.4032\times {10}^{12}}{1.57\times {10}^6\text{ m}}\right)}^{\frac{1}{2}}\\ ={\left(4.07\times {10}^6\right)}^{\frac{1}{2}}\\ =2.02\times {10}^3\text{ m/s}\end{array}$

$\begin{array}{c}{v}_{escape}=2.02\times {10}^3\text{ m/s}\left(\frac{\text{1 }\text{}\text{km}}{1000\text{ m}}\right)\\ =2.02\text{ km/s}\end{array}$

Conclusion:

Therefore, the escape velocity to leave Europa is $2.02\text{ km/s}$.

(b)

To determine

The escape speed from Jupiter at the distance of Europa’s orbit, if a spacecraft landed on Jupiter. Given that the orbit of Europa has a radius of $670900\text{ km}$.

Answer to Problem 13Q

Solution:

$19.28\text{ km/s}$

Explanation of Solution

Given data:

A spacecraft landed on Jupiter’s moon, which moves around Jupiter in an orbit of radius $670900\text{ km}$.

Formula used:

The relation for the escape velocity is as:

${\left(v_{escape}\right)}_{jupiter}={\left(\frac{2Gm}{r}\right)}^{\frac{1}{2}}$

Here, ${\left(v_{escape}\right)}_{jupiter}$ is the escape velocity for Jupiter, G is the universal gravitational constant having value $6.67\times {10}^{-11}\text{ N}\cdot {\text{m}}^{\text{2}}{\text{/kg}}^{\text{2}}$, m is the mass of Jupiter and r is the radius of Jupiter.

Explanation:

The mass of Jupiter is $1.89\times {10}^{27}\text{ kg}$ and the radius of Jupiter is $6.78\times {10}^8\text{ m }$.

Recall the above formula.

${\left(v_{escape}\right)}_{jupiter}={\left(\frac{2Gm}{r}\right)}^{\frac{1}{2}}$

Substitute $1.89\times {10}^{27}\text{ kg}$ for m, $6.78\times {10}^8\text{ m }$ for r and $6.67\times {10}^{-11}\text{ N}\cdot {\text{m}}^{\text{2}}{\text{/kg}}^{\text{2}}$ for G

$\begin{array}{c}{\left(v_{escape}\right)}_{jupiter}={\left(\frac{2\left(6.67\times {10}^{-11}{\text{ Nm}}^{\text{2}}{\text{/kg}}^{\text{2}}\right)\left(1.89\times {10}^{27}\text{ kg}\right)}{6.78\times {10}^8\text{ m}}\right)}^{\frac{1}{2}}\\ ={\left(\frac{2.52\times {10}^{17}}{6.78\times {10}^8\text{ m}}\right)}^{\frac{1}{2}}\\ ={\left(371.86\times {10}^6\right)}^{\frac{1}{2}}\\ =19.28\times {10}^3\text{ m/s}\end{array}$

$\begin{array}{c}{\left(v_{escape}\right)}_{jupiter}=19.28\times {10}^3\text{ m/s}\left(\frac{\text{1 }\text{}\text{km}}{1000\text{ m}}\right)\\ =19.28\text{ km/s}\end{array}$

Conclusion:

Therefore, the escape speed from Jupiter at the distance of Europa’s orbit is $19.28\text{ km/s}$.

(c)

To determine

The explanation that the space craft must leave Europa with a speed greater than the speed, which is calculated in part (a) or part (b), in order to begin its homeward journey if the spacecraft landed on Jupiter’s moon, which revolves around Jupiter in an orbit of radius $670900\text{ km}$. After collecting the sample from the satellite’s surface, the spacecraft prepares its return to Earth.

Answer to Problem 13Q

Solution:

The large escape velocity is required, to cross Jupiter, as compared to Europa’s escape velocity.

Explanation of Solution

Introduction:

Escape velocity: The lowest velocity which is required by the body to escape the gravitational attraction of a particular planet is known as its escape velocity.

Explanation:

The escape velocity for Europa is $2.02\text{ km/s}$, which is calculated in the part (a) and the escape velocity for Jupiter is $19.28\text{ km/s}$, which is calculated in part (b).

From the above calculations, it can be observed that the escape velocity for Europa is less than the escape velocity for Jupiter. If a spacecraft achieves a velocity which is equal to the escape velocity of Europa, then the spacecraft cannot cross the Jovian system. There is requirement of larger velocity than the escape velocity of Europa, to escape the Jovian system for the spacecraft.

Conclusion:

Therefore, for the spacecraft to escape from the Jovian system, an escape velocity larger than that of Europa is required.