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An astronaut's pack weighs 18.8N when she is on earth but only 1.95 N when she...

An astronaut's pack weighs 18.8N when she is on earth but only 1.95 N when she is at the surface of an asteroid. 1)What is the acceleration due to gravity on this asteroid? 2)What is the mass of the pack on the asteroid?
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Answer #1
Concepts and reason

The concepts used to solve this problem are weight and acceleration.

First, calculate the ratio of the weight of the pack on both surfaces that are on earth as well as on asteroid. It will give ratio acceleration of pack on both surfaces. Then multiply the ratio with the acceleration due to gravity of earth that is gg . This will give the value of the acceleration of the pack on the asteroid. Again, divide the weight of the pack on an asteroid by the acceleration of the pack on the asteroid. This will give the mass of pack on an asteroid.

Fundamentals

The relation between mass and weight is,

W=maW = ma

Here mm is the mass of the object and aa is acceleration of the object.

The acceleration aa could be due to earth or any other asteroid on the which the object is situated.

(1)

Ratio of the weights on both surfaces can be calculated as,

WaWE=mamg\frac{{{W_a}}}{{{W_E}}} = \frac{{ma}}{{mg}}

Here Wa{W_a} is the weight on the asteroid, WE{W_E} is weight on earth, mm is mass of pack, aa is acceleration due to an asteroid and gg is acceleration due to gravity.

Rearrange above equation for aa ,

a=gWaWEa = g\frac{{{W_a}}}{{{W_E}}}

Substitute 1.95N1.95{\rm{ N}} for Wa{W_a} , 18.8N18.8{\rm{ N}} for WE{W_E} and 9.8m/s29.8{\rm{ m/}}{{\rm{s}}^2} for gg in the above equation,

a=(9.8m/s2)1.95N18.8N=1.02m/s2\begin{array}{c}\\a = \left( {9.8{\rm{ m/}}{{\rm{s}}^2}} \right)\frac{{1.95{\rm{ N}}}}{{18.8{\rm{ N}}}}\\\\ = 1.02{\rm{ m/}}{{\rm{s}}^2}\\\end{array}

(2)

Wight of the pack on asteroid is related with the mass is as follows,

Wa=ma{W_a} = ma

Rearrange above equation for mm ,

m=Waam = \frac{{{W_a}}}{a}

Substitute 1.95N1.95{\rm{ N}} for Wa{W_a} and 1.02m/s21.02{\rm{ m/}}{{\rm{s}}^2} for aa ,

m=1.95N1.02m/s2=1.92kg\begin{array}{c}\\m = \frac{{1.95{\rm{ N}}}}{{1.02{\rm{ m/}}{{\rm{s}}^2}}}\\\\ = 1.92{\rm{ kg}}\\\end{array}

Ans: Part 1

The magnitude of the acceleration of the pack on the asteroid is 1.02m/s21.02{\rm{ m/}}{{\rm{s}}^2} .

Part 2

The mass of the pack on the asteroid is 1.92kg1.92{\rm{ kg}} .

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