Question

What is the apparent weight of a $\mathbf{75}-\mathbf{kg}$astronaut $\mathbf{2500}\mathbf{km}$from the centre of the Moon in a space vehicle (a) moving at constant velocity and (b) accelerating toward the Moon at $\mathbf{1}\mathbf{.}\mathbf{8}\mathbf{m}\mathbf{/}{\mathbf{s}}^{\mathbf{2}}$? State “direction” in each case?

Short answer

(a) The apparent weight of $75\mathrm{kg}$astronaut $2500\mathrm{km}$from the center of the Moon in a space vehicle moving at constant velocity is $59.25\mathrm{N}$, and the direction is towards the Moon.

(b) The apparent weight of $75\mathrm{kg}$astronaut $2500\mathrm{km}$from the center of the Moon in a space vehicle accelerating towards the Moon at $1.8\mathrm{m}/{\mathrm{s}}^2$is $-75.75\mathrm{N}$, and the direction is away from the Moon.

Step-by-step solution

Step 1. Definition of Velocity, Acceleration, and Acceleration due to gravity.

The rate of change of displacement per unit time is called velocity.

Acceleration is the fraction of the force acting on the object to that of the mass.

The acceleration due to gravity is the term that is produced due to the force of gravity.

The formula for the acceleration due to gravity is,

$g=\frac{GM}{r^2}$

Step 2. Given data

The mass of astronaut $m=75\mathrm{kg}$.

The distance from centre of Moon $r=2500\mathrm{km}=2.5\times {10}^6\mathrm{m}$.

Standard values:

The mass of Moon $M=7.4\times {10}^{22}\mathrm{kg}$.

Step 3. Find the acceleration.

The expression for the Acceleration due to gravity is given as,

$\begin{array}{c}g=\frac{F_{mw}}{M}\\ =\frac{GM}{r^2}\end{array}$

Here, $G$is the gravitational constant and $M$is the mass of the Moon.

Substitute the values in the above equation,

$\begin{array}{c}g=\frac{\left(6.67\times {10}^{-41}\mathrm{N}·{\mathrm{m}}^2/{\mathrm{kg}}^2\right)\left(7.4\times {10}^{-7}\mathrm{kg}\right)}{{\left(2.5\times {10}^4\mathrm{m}\right)}^2}\\ =0.79\mathrm{m}/{\mathrm{s}}^2\end{array}$

Step 4. Calculation of Force when the velocity is constant

Taking a positive direction towards the Moon and the apparent weight is determined by force.

Therefore, the relation between the force and weight becomes,

$-F_N+mg=ma$

For constant velocity, the acceleration becomes zero.

$F_N=mg$

Substitute the values in the above equation,

$\begin{array}{c}{F}_N=(75\mathrm{kg})·\left(0.79\mathrm{m}/{\mathrm{s}}^2\right)\\ =59.25\mathrm{N}\end{array}$

Thus, the apparent weight of $75\mathrm{kg}$astronaut $2500\mathrm{km}$from the center of the Moon in a space vehicle moving at constant velocity is $59.25\mathrm{N}$, and the direction is towards the Moon.

Step 5. Calculation of Force when the acceleration is 1.8 m/s2

The apparent weight when the space vehicle accelerating towards the Moon at $1.8\mathrm{m}/{\mathrm{s}}^2$can be calculated as,

$\begin{array}{c}{F}_N=mg-ma\\ F_N=m(g-a)\end{array}$

Substitute the values in the above equation,

$\begin{array}{c}{F}_N=(75\mathrm{kg})\left(0.79\mathrm{m}/{\mathrm{s}}^2-1.8\mathrm{m}/{\mathrm{s}}^2\right)\\ =-75.75\mathrm{N}\end{array}$

Thus, the apparent weight of $75\mathrm{kg}$astronaut $2500\mathrm{km}$from the center of the Moon in a space vehicle accelerating towards the Moon at $1.8\mathrm{m}/{\mathrm{s}}^2$is $-75.75\mathrm{N}$, and the direction is away from the Moon.