Question

Imagine a landing craft approaching the surface of Callisto, one of Jupiter’s moons. If the engine provides an upward force (thrust) of 3260N, the craft descends at constant speed; if the engine provides only 2200 N, the craft accelerates downward at$\mathbf{0}\mathbf{.}\mathbf{39}\mathbf{m}\mathbf{/}{\mathbf{s}}^{\mathbf{2}}$. (a) What is the weight of the landing craft in the vicinity of Callisto’s surface? (b) What is the mass of the craft? (c) What is the magnitude of the free-fall acceleration near the surface of Callisto?

Short answer

(a) Weight of the landing craft in the vicinity of the Callisto’s surface is 3260N.

(b) Mass of the craft$2.7\times {10}^3\mathrm{kg}$

(b) Magnitude of free fall acceleration near the surface of Callisto $1.2\mathrm{m}/{\mathrm{s}}^2$.

Step-by-step solution

Given information

1. $F_1=3260\mathrm{N}$,
2. $F_2=2200\mathrm{N}$and
3. $a=0.39\mathrm{m}/{\mathrm{s}}^2$
   

Understanding the concepta

The weight of the object is equal to the product of mass and gravitational acceleration. The weight is form of force. Newton’s law states that the force acting on the object is equal to mass times acceleration of the object.

Find weight of the landing craft by using weight equation. We use Newton’s second law to find acceleration.

Formula used

Formulae:

1. W=mg
2. F =ma

Here,W is weight of the object,m is mass of the object,g is gravitational acceleration,F is force anda is acceleration of the object.

(a) Calculate the weight of the craft

To descend at constant speed the force by thrust will be equal to weight of the craft.

We can take downward as positive direction for simplicity; we apply Newton’s second law,

$$\begin{gathered} ma=mg-F_1 \\ m\left(0\right)=mg-3260 \\ mg=3260\mathrm{N} \\ W=3260\mathrm{N} \end{gathered}$$

Weight of the craft is 3260 N.

(b) Calculate the mass of the craft

For second force there is acceleration $a_1=0.39\mathrm{m}/{\mathrm{s}}^2$.

We can apply Newton’s second law,

$$\begin{gathered} m{a}_1=mg-F_2 \\ m=\frac{mg-F_2}{a_1} \\ =\frac{3260\mathrm{N}-2200\mathrm{N}}{0.39\mathrm{m}/{\mathrm{s}}^2} \\ =2.7\times {10}^3\mathrm{kg} \end{gathered}$$

The mass of the craft is $2.7\times {10}^3\mathrm{kg}$.

(c) Calculate the free fall acceleration

Use the weight of the craft to find the free fall acceleration.

$$\begin{gathered} W=mg \\ g=\frac{W}{m} \\ g=\frac{3260\mathrm{N}}{2.7\times {10}^3\mathrm{kg}} \\ =1.2\mathrm{m}/{\mathrm{s}}^2 \end{gathered}$$

Therefore, the free all acceleration is $1.2\mathrm{m}/{\mathrm{s}}^2$.