Chapter 5, Problem 52P

To determine

Part (a)

  $$ m=75\text{kg}$58.8\text{N}$

The apparent weight of an astronaut 2500 km from the center of the Moon in a space vehicle which moves at a constant velocity.

Answer to Problem 52P

Solution:

The apparent weight of the astronaut is and acts towards the center of the Moon.

Explanation of Solution

Given:

Mass of the astronaut,

The orbiting radius of the space vehicle, $r=2500\text{km}$

Formula Used:

Gravitational force:

  $F=G\frac{M_{m}m}{r^2}$

Also, $F=m{g}_m$

Where, $G$ is the gravitational constant

  $M_m$ is the mass of the Moon

m is the mass of the astronaut

r is the distance between M and m

  $g_m$ is the gravitational field of moon

Calculation:

The acceleration due to gravity for Moon at a distance 25000 km from its center is given by

  $m{g}_m=G\frac{M_{m}m}{r^2}$

  $g_m=\left(6.67\times {10}^{-11}\right)\frac{7.342\times {10}^{22}}{{\left(2500\times {10}^3\right)}^2}$

  $g_m=0.784{\text{m/s}}^{\text{2}}$

Now at a constant velocity, the acceleration of the spaceship is zero.

So, the apparent weight of the astronaut is:

  $F_n=m{g}_m$

Substitute the values and solve:

  $F_n=75\times 0.784$

  $F_n=58.8\text{N}$

This is the apparent weight of the astronaut and acts towards the center of the Moon.

Conclusion:

Therefore, the apparent weight of the astronaut is $58.8\text{N}$ and acts towards the center of the Moon.

Part (b)

To determine

The apparent weight of an astronaut 2500 km from the center of the Moon when the space vehicle accelerates at $1.8{\text{m/s}}^{\text{2}}$ .

Answer to Problem 52P

Solution:

Therefore, the apparent weight of the astronaut is $\text{76}\text{.2}\text{N}$ and acts away from the center of the Moon.

Explanation of Solution

Given:

Mass of the astronaut, $m=75\text{kg}$

The orbiting radius of the space vehicle, $r=2500\text{km}$

Formula Used:

Gravitational force:

  $F=G\frac{M_{m}m}{r^2}$

Also, $F=m{g}_m$

Where, $G$ is the gravitational constant

  $M_m$ is the mass of the Moon

m is the mass of the astronaut

r is the distance between M and m

  $g_m$ is the gravitational field of moon

Calculation:

The acceleration due to gravity for Moon at a distance 2500 km from its center is given by

  $m{g}_m=G\frac{M_{m}m}{r^2}$

  $g_m=\left(6.67\times {10}^{-11}\right)\frac{7.342\times {10}^{22}}{{\left(2500\times {10}^3\right)}^2}$

  $g_m=0.784{\text{m/s}}^{\text{2}}$

According to Newton’s second law of motion, $\sum F=ma$

  $F_n=m{g}_m-ma$

  $F_n=m\left(g_m-a\right)$

Substitute the values and solve:

  $F_n=75\left(0.784-1.8\right)$

  $F_n=-76.2\text{N}$

Conclusion:Therefore, the apparent weight of the astronaut is $\text{76}\text{.2}\text{N}$ and acts away from the center of the Moon.