0.2 Force, momentum and impulse (Page 21/35)

Maths test

Page 21 / 35

a_{o} = \frac{F}{m_{o}}

So we substitute equation [link] into Equation [link] , and we find that

a_{o} = \frac{G m_{e}}{r^{2}}

Since it doesn't matter what $m_{o}$ is, this tells us that the acceleration on a body (due to the Earth's gravity) does not depend on the mass of the body. Thus all objects experience the same gravitational acceleration. The force on different bodies will be different but the acceleration will be the same. Due to the fact that this acceleration caused by gravity is the same on all objects we label it differently, instead of using $a$ we use $g$ which we call the gravitational acceleration.

Comparative problems

Comparative problems involve calculation of something in terms of something else that we know. For example, if you weigh 490 N on Earth and the gravitational acceleration on Venus is 0,903 that of the gravitational acceleration on the Earth, then you would weigh 0,903 x 490 N = 442,5 N on Venus.

Principles for answering comparative problems

Write out equations and calculate all quantities for the given situation
Write out all relationships between variable from first and second case
Write out second case
Substitute all first case variables into second case
Write second case in terms of first case

A man has a mass of 70 kg. The planet Zirgon is the same size as the Earth but has twice the mass of the Earth. What would the man weigh on Zirgon, if the gravitational acceleration on Earth is 9,8 m $\cdot$ s $^{- 2}$ ?

Determine what information has been given
The following has been provided:
- the mass of the man, $m$
- the mass of the planet Zirgon ( $m_{Z}$ ) in terms of the mass of the Earth ( $m_{E}$ ), $m_{Z} = 2 m_{E}$
- the radius of the planet Zirgon ( $r_{Z}$ ) in terms of the radius of the Earth ( $r_{E}$ ), $r_{Z} = r_{E}$
Determine how to approach the problem
We are required to determine the man's weight on Zirgon ( $w_{Z}$ ). We can do this by using:

$w = m g = G \frac{m_{1} \cdot m_{2}}{r^{2}}$

to calculate the weight of the man on Earth and then use this value to determine the weight of the man on Zirgon.
Situation on Earth
$\begin{matrix} w_{E} & = & m g_{E} = G \frac{m_{E} \cdot m}{r_{E}^{2}} \\ = & (70 kg) (9, 8 m \cdot s^{- 2}) \\ = & 686 N \end{matrix}$
Situation on Zirgon in terms of situation on Earth
Write the equation for the gravitational force on Zirgon and then substitute the values for $m_{Z}$ and $r_{Z}$ , in terms of the values for the Earth.

$\begin{matrix} w_{Z} = m g_{Z} & = & G \frac{m_{Z} \cdot m}{r_{Z}^{2}} \\ = & G \frac{2 m_{E} \cdot m}{r_{E}^{2}} \\ = & 2 (G \frac{m_{E} \cdot m}{r_{E}^{2}}) \\ = & 2 w_{E} \\ = & 2 (686 N) \\ = & 1 372 N \end{matrix}$
Quote the final answer
The man weighs 1 372 N on Zirgon.

A man has a mass of 70 kg. On the planet Beeble how much will he weigh if Beeble has mass half of that of the Earth and a radius one quarter that of the Earth. Gravitational acceleration on Earth is 9,8 m $\cdot$ s $^{- 2}$ .

Determine what information has been given
The following has been provided:
- the mass of the man on Earth, $m$
- the mass of the planet Beeble ( $m_{B}$ ) in terms of the mass of the Earth ( $m_{E}$ ), $m_{B} = \frac{1}{2} m_{E}$
- the radius of the planet Beeble ( $r_{B}$ ) in terms of the radius of the Earth ( $r_{E}$ ), $r_{B} = \frac{1}{4} r_{E}$
Determine how to approach the problem
We are required to determine the man's weight on Beeble ( $w_{B}$ ). We can do this by using:

$w = m g = G \frac{m_{1} \cdot m_{2}}{r^{2}}$

to calculate the weight of the man on Earth and then use this value to determine the weight of the man on Beeble.
Situation on Earth
$\begin{matrix} w_{E} & = & m g_{E} = G \frac{m_{E} \cdot m}{r_{E}^{2}} \\ = & (70 kg) (9, 8 m \cdot s^{- 2}) \\ = & 686 N \end{matrix}$
Situation on Beeble in terms of situation on Earth
Write the equation for the gravitational force on Beeble and then substitute the values for $m_{B}$ and $r_{B}$ , in terms of the values for the Earth.

$\begin{matrix} w_{B} = m g_{B} & = & G \frac{m_{B} \cdot m}{r_{B}^{2}} \\ = & G \frac{\frac{1}{2} m_{E} \cdot m}{{(\frac{1}{4} r_{E})}^{2}} \\ = & 8 (G \frac{m_{E} \cdot m}{r_{E}^{2}}) \\ = & 8 w_{E} \\ = & 8 (686 N) \\ = & 5 488 N \end{matrix}$
Quote the final answer
The man weighs 5 488 N on Beeble.

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Source: OpenStax, Maths test. OpenStax CNX. Feb 09, 2011 Download for free at http://cnx.org/content/col11236/1.2

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