91Ó°ÊÓ

• The density of the Martian atmosphere is,${\mathrm{ÒÏ}}_m=0.0154kg/m^3$.
• The density of Earth’s atmosphere is, ${\mathrm{ÒÏ}}_a=1.20kg/m^3$.

• The mass of a balloon per square meter is, $m=5g/m^2$.

The density of an object in the given space is the mass occupied by the object in the unit volume. It varies with the temperature of the object.

The volume of the balloon is given as:

$V=\frac{4}{3}{\mathrm{Ï€°ù}}^3$ (1)

Here r is the radius of the sphere.

The total mass of the balloon is given as:

$\begin{array}{r}M=m\left(4\mathrm{Ï€}{r}^2\right)\\ {\mathrm{ÒÏ}}_mâ‹\dots V=m\left(4\mathrm{Ï€}{r}^2\right)\\ {\mathrm{ÒÏ}}_m\left(\frac{4}{3}\mathrm{Ï€}{r}^3\right)=m\left(4\mathrm{Ï€}{r}^2\right)\\ r=\frac{3m}{{\mathrm{ÒÏ}}_m}\end{array}$ (2)

Substitute all the values in the above equation to find the radius of each balloon, and we get,

$\begin{array}{r}r=\frac{3\left(5\mathrm{g}/{\mathrm{m}}^2\right)\left(\frac{1\mathrm{kg}}{1000\mathrm{g}}\right)}{0.0154\mathrm{kg}/{\mathrm{m}}^3}\\ r=0.974\mathrm{m}\end{array}$

From equation (2), we can calculate the mass of the balloon as:

$\begin{array}{r}M=\left(5\mathrm{g}/{\mathrm{m}}^2\right)\left(\frac{1\mathrm{kg}}{1000\mathrm{g}}\right)\left(4\mathrm{π}(0.974\mathrm{m}{)}^2\right)\\ M=5.96×{10}^{−2}\mathrm{kg}\end{array}$

Therefore, the radius and mass of the balloon is 0.974 m and $5.96×{10}^{-2}kg$.

Substitute all the values in equation (1) to find the volume of the balloon; we get,

$\begin{array}{r}V=\frac{4}{3}\mathrm{Ï€}(0.974\mathrm{m}{)}^3\\ V=3.87{\mathrm{m}}^3\end{array}$

The initial acceleration of a balloon on Earth can be calculated as,

$a=\frac{\left({\mathrm{ÒÏ}}_a∨g−mg\right)}{m}$

Substitute all the values in the above equation, and we get,

$\begin{array}{r}a=\frac{\left[\left(1.20\mathrm{kg}/{\mathrm{m}}^3\right)\left(3.87{\mathrm{m}}^3\right)\left(9.81\mathrm{m}/{\mathrm{s}}^2\right)−\left(5.96×{10}^{−2}\mathrm{kg}\right)\left(9.81\mathrm{m}/{\mathrm{s}}^2\right)\right]}{\left(5.96×{10}^{−2}\mathrm{kg}\right)}\\ a=755\mathrm{m}/{\mathrm{s}}^2\end{array}$

Therefore, the initial acceleration of the balloon on Earth is 755$m/s^2$ , and the balloon will go up.

The volume of the total system on Mars is given as:

$V_m=\frac{4}{3}\mathrm{Ï€}{r}_m^3$

Here, $r_m$ is the radius of the balloon on Mars, and its value is five times the radius of the balloon on Earth.

Substitute all the values in the above equation, and we get,

$\begin{array}{r}{V}_m=\frac{4}{3}\mathrm{Ï€}(5r{)}^3\\ V_m=\frac{4}{3}\mathrm{Ï€}\left(5\right(0.974\mathrm{m}){)}^3\\ V_m=438.81{\mathrm{m}}^3\end{array}$

The mass of the total system on Mars is given as:

$M_m=4\mathrm{Ï€}{r}_m^{2}m$

Substitute all the values in the above equation, and we get,

$\begin{array}{r}{M}_m=4\mathrm{Ï€}(5r{)}^2\\ M_m=4\mathrm{Ï€}\left(5\right(0.974\mathrm{m}){)}^2\left(5\mathrm{g}/{\mathrm{m}}^2\right)\left(\frac{1\mathrm{kg}}{1000\mathrm{g}}\right)\\ M_m=1.490\mathrm{kg}\end{array}$

The mass of the instrument package is given as:

$m_p={\mathrm{ÒÏ}}_m{V}_m−M_m$

Substitute all the values in the above equation, and we get,

$\begin{array}{r}{m}_p=\left(0.0154\mathrm{kg}/{\mathrm{m}}^3\right)\left(438.81{\mathrm{m}}^3\right)−1.490\mathrm{kg}\\ m_p=5.27\mathrm{kg}\end{array}$

Therefore, the mass of the instrument package is 5.27 kg .