Mechanics: Identify and apply principles of force

# Unit 3: Mass and weight

Linda Pretorius

### Unit outcomes

By the end of this unit you will be able to:

- Define gravitational acceleration:
*g* - Describe weight.
- Differentiate between mass (
*m*) and weight (*F*_{gravitation}). - Calculate weight.

## What you should know

Before you start this unit, make sure you can:

- Distinguish between a vector and a scalar. Subject Outcome 2.1, Unit 3.
- Define a force. Subject Outcome 2.2, Unit 1.
- Say how gravity acts on an object. Subject Outcome 2.2, Unit 1.
- Define acceleration. Subject Outcome 2.1, Unit 1.
- Use equations of motion to determine acceleration. Subject Outcome 2.1, Unit 2.

## Introduction

In this unit^{[1]} you will learn that in physics there is a difference between mass and weight. Your understanding of forces so far will help you to think of weight in terms of gravity, and you will learn how to calculate weight depending on what the is where you are.

# Gravity

is quite an intuitive concept. We realise from a young age that things fall to the ground when we drop them from a height. In the two previous units, you have defined gravity more formally as a non-contact force and included it in your force diagrams as a downward force.

Gravity is the force that arises between objects because of their mass. It is always an attractive force and acts at a distance. We experience the effect of the Earth’s gravity as notable, because the Earth’s mass is so much bigger than any other object on Earth.

## Newton’s universal law of gravitation

Sir Isaac Newton was the first scientist to define the gravitational force, and to show that it affects all bodies with mass. Newton’s universal law of gravitation states that every particle in the universe attracts every other particle. The force is directly proportional to the product of their masses and inversely proportional to the square of the distance between them. The magnitude of the gravitational force between two objects is shown by this formula:

[latex]\scriptsize F=\text{G}\displaystyle \frac{{{{m}_{1}}{{m}_{2}}}}{{{{d}^{2}}}}[/latex]

With:

*F* = the magnitude of the gravitational force in newtons (N)

G = the gravitational constant [latex]\scriptsize 6.67\times {{10}^{{-11}}}\text{ N}\cdot {{\text{m}}^{2}}\cdot \text{k}{{\text{g}}^{{-2}}}[/latex]

*m*_{1} = the mass of the first object in kilograms (kg)

*m*_{2} = the mass of the second object in kilograms (kg)

*d* = the distance between the objects in metres (m).

For large objects, we use the distance from the centre of the object(s) to do the calculation. This is especially important when dealing with very large objects such as planets. The distance from the centre of the planet and from the surface of the planet can differ by a large amount.

### Note

When doing calculations involving planets, check carefully whether the diameter or radius is given as an indication of distance.

Let’s look at an example of using Newton’s universal law of gravitation in a calculation.

- Consider a person with a mass of [latex]\scriptsize 80\text{ kg}[/latex] standing [latex]\scriptsize 10\text{ m}[/latex] from someone else with a mass of [latex]\scriptsize \text{65 kg}[/latex]. The attractive gravitational force between them would be:

[latex]\scriptsize \begin{align*}F & =\text{G}\displaystyle \frac{{{{m}_{1}}{{m}_{2}}}}{{{{d}^{2}}}}\\ & =(6.67\times {{10}^{{-11}}})\left( {\displaystyle \frac{{(80)(65)}}{{{{{(10)}}^{2}}}}} \right)\\ & =3.47\times {{10}^{{-9}}}\text{ N}\end{align*}[/latex]

. - If the two people were only [latex]\scriptsize 1\text{ m}[/latex] apart, then the force is:

[latex]\scriptsize \begin{align*}F & =\text{G}\displaystyle \frac{{{{m}_{1}}{{m}_{2}}}}{{{{d}^{2}}}}\\ & =(6.67\times {{10}^{{-11}}})\left( {\displaystyle \frac{{(80)(65)}}{{{{{(1)}}^{2}}}}} \right)\\ & =3.47\times {{10}^{{-7}}}\text{ N}\end{align*}[/latex]

. - Now consider the gravitational force between the Earth and the Moon. The mass of the Earth is [latex]\scriptsize 5.98\times {{10}^{{24}}}\text{ kg}[/latex], the mass of the Moon is [latex]\scriptsize 7.35\times {{10}^{{22}}}\text{ kg}[/latex] and the Earth and Moon are apart. The gravitational force between the Earth and Moon is:

[latex]\scriptsize \begin{align*}F & =\text{G}\displaystyle \frac{{{{m}_{1}}{{m}_{2}}}}{{{{d}^{2}}}}\\ & =(6.67\times {{10}^{{-11}}})\left( {\displaystyle \frac{{(5.98\times {{{10}}^{{24}}})(7.35\times {{{10}}^{{22}}})}}{{{{{(0.38\times {{{10}}^{9}})}}^{2}}}}} \right)\\ & =2.03\times {{10}^{{20}}}\text{ N}\end{align*}[/latex]

The equation mathematically confirms what we see through observation:

- The greater the distance between the two interacting objects, the smaller the gravitational force between them.
- The bigger the masses of the two interacting objects, the bigger the gravitational force will be.

### Note

Gravitational force is a vector. We use Newton’s law of universal gravitation to determine the magnitude of the force and then analyse the problem to determine the direction.

### Note

If you want to explore how the masses of two interacting objects and the distance between them affect their gravitational force, use this simulation:

## Gravitational acceleration

You already know that a non-zero resultant force can change the speed or direction in which an object moves. You also know from your study of motion in one dimension that a change to an object’s speed or direction is called acceleration. Therefore, force can cause an acceleration of an object.

The acceleration caused by the force of gravity on a free-falling object is called gravitational acceleration. It is represented by the symbol *g*, and has a constant value of [latex]\scriptsize 9.8\text{ m}\cdot {{\text{s}}^{{-2}}}[/latex] on Earth.

gravitational acceleration (*g*) on Earth = [latex]\scriptsize 9.8\text{ m}\cdot {{\text{s}}^{{-2}}}[/latex].

In Activity 3.1 you will investigate the value of gravitational acceleration experimentally.

### Activity 3.1: Determine *g* by experiment

**Time required:** 10 minutes

**What you need:**

- a smartphone with a QR scanner
- software to view video files frame by frame (e.g. Windows Media Player)
- calculator, pen and notebook

**What to do:**

- Use this QR code to download the video file.

- Watch the video once.
- Find the point in the video shortly before the ball is released. Then scroll forward frame by frame to the point where the ball is released. You can do this in Windows Media Player by holding down Ctrl and pressing the play button repeatedly.
- Count the number of frames from the exact point when the ball is released to where it hits the ground. You may have to do this a few times to make sure you have counted accurately.
- Calculate the time it took for the ball to reach the ground. The video was recorded at [latex]\scriptsize 30[/latex] frames per second.
- Use the result and the equation [latex]\scriptsize s=ut+\displaystyle \frac{1}{2}a{{t}^{2}}[/latex] to determine
*g*. The ball was released from a height of [latex]\scriptsize 2.1\text{ m}[/latex]. - Discuss the accuracy of the result and how the experiment could be improved.

**What did you find?**

It takes [latex]\scriptsize 20[/latex] frames from when the ball is released to when it hits the ground. So, at [latex]\scriptsize 30[/latex] frames per second, it took [latex]\scriptsize \displaystyle \frac{{20}}{{30}}=0.67[/latex] seconds for the ball to reach the ground.

In the equation [latex]\scriptsize s=ut+\displaystyle \frac{1}{2}a{{t}^{2}}[/latex], the term [latex]\scriptsize ut=0[/latex], because the ball falls from rest; in other words, the initial velocity is zero. Acceleration, *a*, can be represented by the symbol *g* for gravitational acceleration. The equation can then be rearranged as follows to calculate *g*:

[latex]\scriptsize \begin{align*}s & =ut+\displaystyle \frac{1}{2}a{{t}^{2}}\\g & =\displaystyle \frac{{2s}}{{{{t}^{2}}}}\\ & =\displaystyle \frac{{2(2.1)}}{{{{{(0.67)}}^{2}}}}\\ & =9.4\text{ m}\cdot {{\text{s}}^{{-2}}}\end{align*}[/latex]

This is only an approximation of the value of *g*, but is fairly close to the actual value of [latex]\scriptsize 9.8\text{ m}\cdot {{\text{s}}^{{-2}}}[/latex]. The margin of uncertainty in the time measurement is [latex]\scriptsize 0.03\text{ s}[/latex], which contributes to the value not being exactly [latex]\scriptsize 9.8\text{ m}\cdot {{\text{s}}^{{-2}}}[/latex]. Accuracy could have been improved if the recording were made at a higher frame rate (e.g. 60 frames per second), if the height had been controlled more precisely and if there was no air resistance. However, despite these limitations, the experiment yields a reasonable approximation of *g*. Why not try the experiment yourself?

### Note

For an alternative approach to this experiment, using more refined laboratory equipment, watch the video called ‘Experiment to determine *g*’ (duration: 10:37).

Although the magnitude of *g* is different on different planets, it is always constant on a particular planet. That means all experience the same gravitational acceleration ([latex]\scriptsize 9.8\text{ m}\cdot {{\text{s}}^{{-2}}}[/latex] on Earth), regardless of their mass. A crate of gold will experience the same gravitational acceleration as a crate of cotton wool.

### Did you know?

The Italian scientist and mathematician Galileo Galilei had already predicted the constant gravitational acceleration of free-falling bodies in the late 16th century. To see this yourself, watch the videos of this experiment on Earth (duration: 04:41 minutes) and on the Moon (duration: 00:51 minutes):

### Activity 3.2: Research the value of g on the Moon and Mars

**Time required:** 10 minutes

**What you need:**

- a phone with a QR scanner
- an internet connection
- calculator, pen and notebook

**What to do:**

- Scan the QR code and download the fact sheet:

- Find the gravitational acceleration (
*g*) on Mars and the Moon. - Give the values relative to that of the Earth’s gravitational acceleration.
- Challenge question: How do you think you can check whether NASA’s measurements are correct?

Hint:

Newton’s second law (which you will learn about in Level 3) is given by: [latex]\scriptsize F=ma[/latex].

You also know that Newton’s universal law of gravitation is given by: [latex]\scriptsize F=\text{G}\displaystyle \frac{{{{m}_{1}}{{m}_{2}}}}{{{{d}^{2}}}}[/latex]

**What did you find?**

- Gravitational acceleration on Mars is [latex]\scriptsize 3.7\text{ m}\cdot {{\text{s}}^{{-2}}}[/latex]. It is [latex]\scriptsize 0.38[/latex] times of that on Earth.
- Gravitational acceleration on the Moon is [latex]\scriptsize 1.6\text{ m}\cdot {{\text{s}}^{{-2}}}[/latex]. It is [latex]\scriptsize 0.16[/latex] times of that on Earth.
- Newton’s second law of motion states that the force experienced by an object is equal to the product of the object’s mass and its acceleration. We can therefore replace the term
*F*in the equation of the universal law of gravitation:

[latex]\scriptsize {{m}_{{\text{Earth}}}}a=\text{G}\displaystyle \frac{{{{m}_{{\text{moon}}}}{{m}_{{\text{Earth}}}}}}{{{{d}_{{\text{moon}}}}^{2}}}[/latex]

The mass of the Earth cancels out on either side of the equation and we can write the acceleration as*g*(for gravitational acceleration), rather than*a*. We know from the information sheet that the mass (m) of the Moon is [latex]\scriptsize 0.073\times {{10}^{{24}}}\text{ kg}[/latex] and that its radius (*d*) is [latex]\scriptsize 3.475\times {{10}^{6}}\text{ m }\div \text{2 =1}\text{.737}\times \text{1}{{\text{0}}^{6}}\text{ m}[/latex].

.

The equation then becomes:

[latex]\scriptsize \begin{align*}g & =\text{G}\displaystyle \frac{{{{m}_{{\text{moon}}}}}}{{{{d}^{2}}_{{\text{moon}}}}}\\ & =\displaystyle \frac{{(6.67\times {{{10}}^{{-11}}})(0.073\times {{{10}}^{{24}}})}}{{{{{(1.737\times {{{10}}^{6}})}}^{2}}}}\\ & =1.6\text{ m}\cdot {{\text{s}}^{{-2}}}\end{align*}[/latex]

We can do the same calculation with the information about Mars, and find that [latex]\scriptsize {{g}_{{\text{Mars}}}}=3.7\text{ m}\cdot {{\text{s}}^{{-2}}}[/latex]

# Weight and mass

and are not the same, although we tend to use the terms to mean the same thing in everyday language.

- Mass is a measurement of how much matter an object consists of; it is a scalar quantity. Mass is measured in kilograms (kg).
- Weight is a measurement of how hard gravity is pulling on that object; it is a vector quantity. Weight is therefore simply
*F*_{gravitation}and is measured in newtons (N).

Your **mass** is the same wherever you are – on Earth, on the Moon, on Mars or floating in space – because the amount of matter you are made of does not change. However, your **weight** depends on how strong the gravitational force is that is acting on you at a particular moment. So, you would weigh less on the Moon and on Mars than on Earth, and in space you would weigh almost nothing at all.

### Take note!

When you stand on a scale, you are trying measure how much of you there is. People who are trying to reduce their mass hope to see the reading on the scale get smaller, despite talking about losing weight. If they are successful, their weight will decrease, yes, but it is because their mass is decreasing. A scale therefore uses the person’s weight to determine their mass.

## Calculating weight

We use either one of the following two equations to calculate weight, depending on what information we have available.

[latex]\scriptsize W={{F}_{{\text{gravitation}}}}=mg[/latex]

[latex]\scriptsize {{F}_{{\text{gravitation}}}}=\text{G}\displaystyle \frac{{{{m}_{1}}{{m}_{2}}}}{{{{d}^{2}}}}[/latex]

### Example 3.1

Calculate the weight of a hammer with a mass of [latex]\scriptsize 675\text{ g}[/latex] on Jupiter. Jupiter has a mass of [latex]\scriptsize 1.898\times {{10}^{{27}}}\text{ kg}[/latex] and a diameter of [latex]\scriptsize 142\ 984\text{ km}[/latex].

*Solution*

First analyse the given information:

- The gravitational acceleration of Jupiter is not given, only the masses of objects and the size of Jupiter. It would therefore be best to use the equation [latex]\scriptsize {{F}_{{\text{gravitation}}}}=\text{G}\displaystyle \frac{{{{m}_{1}}{{m}_{2}}}}{{{{d}^{2}}}}[/latex].
- The diameter of Jupiter is given. However, we know that we have to use the distance between the centres of the objects to do the calculation and therefore the radius has to be used: [latex]\scriptsize \displaystyle \frac{{142\ 982}}{2}=71\ 491\text{ km}[/latex]. Remember that this has to be expressed in metres, so [latex]\scriptsize \text{71}\,\text{492 km = 7}\text{.1492 }\times \text{1}{{\text{0}}^{7}}\text{ m}[/latex]
- All masses should be expressed in kilograms, so the mass of the hammer has to be converted: [latex]\scriptsize 675\text{ g = 0}\text{.675 kg}[/latex].

Now do the calculation:

[latex]\scriptsize \begin{align*}W & ={{F}_{{\text{gravitation}}}}=\text{G}\displaystyle \frac{{{{m}_{1}}{{m}_{2}}}}{{{{d}^{2}}}}\\ & =(6.67\times {{10}^{{-11}}})\left( {\displaystyle \frac{{(0.675)(1.898\times {{{10}}^{{27}}})}}{{{{{(7.1492\times {{{10}}^{7}})}}^{2}}}}} \right)\\ & =16.7\text{ N}\end{align*}[/latex]

### Example 3.2

Calculate the weight of a hammer with a mass of [latex]\scriptsize 675\text{ g}[/latex] on Jupiter. The gravitational acceleration on Jupiter is [latex]\scriptsize 24.8\text{ m}\cdot {{\text{s}}^{{-2}}}[/latex].

*Solution*

As the gravitational acceleration is given, we can use the equation [latex]\scriptsize W={{F}_{{\text{gravitation}}}}=mg[/latex]. Remember to convert the mass of the hammer to kilograms.

[latex]\scriptsize \begin{align*}W & ={{F}_{{\text{gravitation}}}}=mg\\ & =0.675\times 24.8\\ & =16.7\text{ N}\end{align*}[/latex]

The weight is the same as determined using the equation for the universal law of gravitation.

### Exercise 3.1

*Questions 1, 2 and 4 were sourced from or based on questions in Siyavula Physical Science Gr 11 Learner’s Book, pp. 114–118 and 133 released under a CC-BY licence at https://www.siyavula.com/read/science/grade-11/newtons-laws/02-newtons-laws-04 and https://www.siyavula.com/read/science/grade-11/newtons-laws/02-newtons-laws-06, respectively.*

- Jojo has a mass of [latex]\scriptsize 87.5\text{ kg}[/latex]. What is his weight on Mercury, which has a gravitational acceleration of [latex]\scriptsize 3.70\text{ m}\cdot {{\text{s}}^{{-2}}}[/latex]?
- If object 1 has a weight of [latex]\scriptsize 1.78\times {{10}^{{3\text{ }}}}\text{N}[/latex] on Neptune and object 2 has a weight of [latex]\scriptsize 3.63\times {{10}^{{\text{5 }}}}\text{N}[/latex] on Mars, which has the greater mass? The gravitational acceleration on Neptune is [latex]\scriptsize 11.15\text{ m}\cdot {{\text{s}}^{{-2}}}[/latex], while it is [latex]\scriptsize 3.71\text{ m}\cdot {{\text{s}}^{{-2}}}[/latex] on Mars.
- The gravitational acceleration on the Moon is about a sixth of that on Earth. What would the approximate weight be of a [latex]\scriptsize 2.5\text{ kg}[/latex] packet of sugar on the Moon?
- A satellite circles around the Earth at a height where the gravitational force is a factor 4 less than at the surface of the Earth. If the Earth’s radius is [latex]\scriptsize R[/latex], what would the height of the satellite be above the surface?
- [latex]\scriptsize R[/latex]
- [latex]\scriptsize 2R[/latex]
- [latex]\scriptsize 4R[/latex]
- [latex]\scriptsize 16R[/latex]

- Challenge question: Your mass is [latex]\scriptsize 60\text{ kg}[/latex] in Paris at ground level. How much less would you weigh after taking a lift to the top of the Eiffel Tower, which is [latex]\scriptsize 405\text{ m}[/latex] high? Assume the Earth’s mass is [latex]\scriptsize 6\times {{10}^{{24}}}\text{ kg}[/latex] and the Earth’s radius is [latex]\scriptsize 6400\text{ km}[/latex].

The full solutions are at the end of the unit.

## Summary

In this unit you have learnt the following:

- Gravity is a force that arises between objects because of their mass.
- Gravity acts at a distance and is always an attractive force.
- Newton’s universal law of gravity states that every particle in the universe attracts every other particle:
- the greater the distance between the two interacting objects, the smaller the gravitational force between them
- the bigger the masses of the two interacting objects, the bigger the gravitational force will be.

- The magnitude of the gravitational force between two objects is given by [latex]\scriptsize F=\text{G}\displaystyle \frac{{{{m}_{1}}{{m}_{2}}}}{{{{d}^{2}}}}[/latex].
- The gravitational constant is [latex]\scriptsize 6.67\times {{10}^{{-11}}}\text{ N}\cdot {{\text{m}}^{2}}\cdot \text{k}{{\text{g}}^{{-2}}}[/latex].
- The acceleration caused by the force of gravity on a free-falling object is called gravitational acceleration,
*g*. - The gravitational acceleration on Earth is [latex]\scriptsize 9.8\text{ m}\cdot {{\text{s}}^{{-2}}}[/latex]. It is constant for all objects, independent of mass.
- Although the magnitude of
*g*is different on different planets, it is always constant on a particular planet. - Mass is a measurement of how much matter is in an object. It is measured in kilograms (kg).
- Weight is a measurement of how hard gravity is pulling on an object. Weight is measured in newtons (N).
- Weight (Fgravitation) is the product of an object’s mass and gravitational acceleration:

[latex]\scriptsize W={{F}_{{\text{gravitation}}}}=mg[/latex]

# Unit 3: Assessment

#### Suggested time to complete: 25 minutes

*Questions 1 and 2 were sourced from Siyavula Physical Science Gr 11 Learner’s Book, pp. 117–119 and pp. 132–133, released under a CC-BY licence at https://www.siyavula.com/read/science/grade-11/newtons-laws/02-newtons-laws-04 and https://www.siyavula.com/read/science/grade-11/newtons-laws/02-newtons-laws-06, respectively.*

For questions 1–3, choose the correct answer.

- Two objects of mass [latex]\scriptsize 2X[/latex] and [latex]\scriptsize 3X[/latex] respectively, where [latex]\scriptsize X[/latex] is an unknown quantity, exert a force [latex]\scriptsize F[/latex] on each other when they are a certain distance apart. What will be the force between two objects situated the same distance apart but having a mass of [latex]\scriptsize 5X[/latex] and [latex]\scriptsize 6X[/latex], respectively?
- [latex]\scriptsize 0.2F[/latex]
- [latex]\scriptsize 1.2F[/latex]
- [latex]\scriptsize 2.2F[/latex]
- [latex]\scriptsize 5F[/latex]

- As the distance of an object above the surface of the Earth is greatly increased, the weight of the object would:
- increase.
- decrease.
- increase and then suddenly decrease.
- remain the same.

- A satellite experiences a force [latex]\scriptsize F[/latex] when at the surface of the Earth. What will be the force on the satellite if it orbits at a height equal to the diameter of the Earth?
- [latex]\scriptsize \displaystyle \frac{1}{F}[/latex]
- [latex]\scriptsize \displaystyle \frac{1}{2}F[/latex]
- [latex]\scriptsize \displaystyle \frac{1}{3}F[/latex]
- [latex]\scriptsize \displaystyle \frac{1}{9}F[/latex]

- Read each of the following statements and say whether you agree or disagree. Give reasons for your answers and rewrite the statement if necessary:
- The gravitational acceleration g is a universal constant.
- The weight of an object is independent of its mass.
- [latex]\scriptsize G[/latex] is dependent on the mass of the object that is being accelerated.

- Halley’s comet, of approximate mass [latex]\scriptsize 1\times {{10}^{{15}}}\text{ kg}[/latex], was [latex]\scriptsize 1.3\times {{10}^{8}}\text{ km}[/latex] from the Earth at its point of closest approach during its last sighting in 1986.
- Name the force through which the Earth and the comet interact.
- Does the acceleration of the comet increase, decrease or remain the same as it moves closer to the Earth? Explain.
- If the mass of the Earth is [latex]\scriptsize 6\times {{10}^{{24}}}\text{ kg}[/latex], calculate the magnitude of the force exerted by the Earth on Halley’s comet at its point of closest approach.

- A man has a mass of [latex]\scriptsize 70\ \text{kg}[/latex] on Earth. He is walking on a new planet that has a mass four times that of the Earth and the radius is the same as that of the Earth ([latex]\scriptsize {{m}_{\text{E}}}=6\times {{10}^{{24}}}\text{ kg}[/latex]; [latex]\scriptsize {{r}_{\text{E}}}=6\times {{10}^{6}}\text{ m}[/latex]).
- Calculate the force between the man and the Earth.
- Would his weight be bigger or smaller on the new planet? Explain your reasoning.
- Calculate the man’s weight on the new planet.

- An astronaut weighs [latex]\scriptsize 750\text{ N}[/latex] on the surface of the Earth. What will his mass be on Saturn, which is [latex]\scriptsize 100[/latex] times more massive than Earth and has a radius [latex]\scriptsize 5[/latex] times that of Earth?

The full solutions are at the end of the unit.

# Unit 3: Solutions

### Exercise 3.1

- .

[latex]\scriptsize \begin{align*}{{F}_{{\text{gravitation}}}} & =W=mg\\ & =(87.5)(3.70)\\ & =323.75\text{ N}\end{align*}[/latex] - The mass of object 1, which is on Neptune, can be calculated as follows:

[latex]\scriptsize \begin{align*}{{W}_{1}} & ={{m}_{1}}g\\\therefore {{m}_{1}} & =\displaystyle \frac{{{{W}_{1}}}}{g}\\ & =\displaystyle \frac{{1.78\times {{{10}}^{3}}}}{{11.15}}\\ & =159.6\text{ kg}\end{align*}[/latex]

.

The mass of object 2, which is on Mars, can be calculated as follows:

[latex]\scriptsize \begin{align*}{{W}_{2}} & ={{m}_{2}}g\\\therefore {{m}_{2}} & =\displaystyle \frac{{{{W}_{2}}}}{g}\\ & =\displaystyle \frac{{3.63\times {{{10}}^{5}}}}{{3.71}}\\ & =9.78\times {{10}^{4}}\text{ kg}\end{align*}[/latex]

.

Object 2 therefore has a greater mass. - We know that weight is directly proportional to gravitational acceleration. Therefore, if the gravitational acceleration on the Moon is about a sixth of that on Earth, we know the weight of the packet of sugar would be reduced by the same factor.

We can calculate the weight of the packet on Earth as:

[latex]\scriptsize \begin{align*}W & =mg\\ & =(2.5)(9.8)\\ & =24.5\text{ N}\end{align*}[/latex]

.

The weight on the Moon would then be approximately [latex]\scriptsize \displaystyle \frac{{24.5}}{6}=4.1\text{ N}[/latex]. - The correct answer is option A:
*R*.

We know that gravitational force is inversely proportional to the square of the distance between the**centres of two objects**. If the gravitational force is [latex]\scriptsize \displaystyle \frac{1}{4}[/latex] at the height where the satellite is orbiting, it means the distance between the centres of the object must be [latex]\scriptsize 2[/latex] times that of the radius of the Earth.

However, the question asks us to calculate the height at which the satellite is orbiting relative to the**surface**of the Earth. Therefore: [latex]\scriptsize 2R-R=R[/latex]. - We start with your weight on the surface of the Earth. The gravitational acceleration at the surface of the Earth is [latex]\scriptsize 9.8\text{ m}\cdot {{\text{s}}^{{-2}}}[/latex], and so your weight is:

[latex]\scriptsize \begin{align*}{{F}_{{\text{gravitation}}}} & =W=mg\\ & =(60)(9.8)\\ & =588\text{ N}\end{align*}[/latex]

.

We know that [latex]\scriptsize {{F}_{{\text{gravitation}}}}=\text{G}\displaystyle \frac{{{{m}_{1}}{{m}_{2}}}}{{{{d}^{2}}}}[/latex] and that [latex]\scriptsize {{F}_{{\text{gravitation}}}}=W=mg[/latex].

Let’s define your mass as*m*_{1}and the mass of the Earth as*m*_{2}. Then we can write:

[latex]\scriptsize {{m}_{1}}g=\text{G}\displaystyle \frac{{{{m}_{1}}{{m}_{2}}}}{{{{d}^{2}}}}[/latex]. The value of*m*_{1}cancels out on either side of the equation, to give the acceleration at the top of the Eiffel Tower as [latex]\scriptsize {{g}_{{\text{Eiffel}}}}=\text{G}\displaystyle \frac{{{{m}_{2}}}}{{{{d}^{2}}}}[/latex]:

[latex]\scriptsize \begin{align*}{{g}_{{\text{Eiffel}}}} & =\text{G}\displaystyle \frac{{{{m}_{2}}}}{{{{d}^{2}}}}\\ & =\displaystyle \frac{{(6.67\times {{{10}}^{{-11}}})(6.0\times {{{10}}^{{24}}})}}{{(6400\times {{10}^{3}}+405)}^{2}}\\ & =9.77\text{ m}\cdot {{\text{s}}^{{-2}}}\end{align*}[/latex]

.

Your weight at the top of the Eiffel Tower would then be:

[latex]\scriptsize \begin{align*}{{W}_{{\text{Eiffel}}}} & =m{{g}_{{\text{Eiffel}}}}\\ & =(60)(9.77)\\ & =586.2\text{ N}\end{align*}[/latex]

.

You would therefore weigh [latex]\scriptsize 1.8\text{ N}[/latex] less at the top of the Eiffel Tower than on the ground.

### Unit 3: Assessment

- D

We can write an expression for the force in each case in terms of the distance.

So, for force 1:

[latex]\scriptsize \begin{align*}{{F}_{1}} & =\text{G}\displaystyle \frac{{{{m}_{1}}{{m}_{2}}}}{{{{d}^{2}}}}\\{{d}^{2}} & =\text{G}\displaystyle \frac{{(2X)(3X)}}{{{{F}_{1}}}}\end{align*}[/latex]

.

For force 2:

[latex]\scriptsize \begin{align*}{{F}_{2}} & =\text{G}\displaystyle \frac{{{{m}_{1}}{{m}_{2}}}}{{{{d}^{2}}}}\\{{d}^{2}} & =\text{G}\displaystyle \frac{{(5X)(6X)}}{{{{F}_{2}}}}\end{align*}[/latex]

.

Since the distance is equal, the square of the distance is also equal and so we can equate these two and find the force.

[latex]\scriptsize \begin{align*}\text{G}\displaystyle \frac{{(2X)(3X)}}{{{{F}_{1}}}} & =\text{G}\displaystyle \frac{{(5X)(6X)}}{{{{F}_{2}}}}\\\displaystyle \frac{{6{{X}^{2}}}}{{{{F}_{1}}}} & =\displaystyle \frac{{30{{X}^{2}}}}{{{{F}_{2}}}}\\\therefore {{F}_{2}} & =5{{F}_{1}}\end{align*}[/latex]

.

An alternative approach would be to look at the proportional increase in mass. The mass of object 1 increases [latex]\scriptsize 2.5[/latex] times, and the mass of object 2 increases [latex]\scriptsize 2[/latex] times. That means that the total mass has increased [latex]\scriptsize 5[/latex] times. Because we know force is directly proportional to the product of the masses, it follows that the force should also increase by a factor [latex]\scriptsize 5[/latex]. - B. The weight would decrease.

The distance is inversely proportional to the weight (or force of gravity) and so, as the distance increases, the weight decreases. - D: [latex]\scriptsize \displaystyle \frac{1}{9}F[/latex]

The diameter of the Earth is twice the radius. This means that the distance to the satellite would be the radius of the Earth plus twice the radius of the Earth, a factor 3 increase. Therefore, [latex]\scriptsize \displaystyle \frac{1}{9}[/latex]. - .
- Disagree. The value of
*g*is constant on a specific planet (e.g. Earth), but it is not the same throughout the universe. - Disagree. Weight is related to mass via the gravitational acceleration.
- Disagree.
*G*is a universal constant: it is the same value throughout the universe.

- Disagree. The value of
- .
- Newton’s law of universal gravitation
- The gravitational force the comet experiences increases because it is inversely proportional to the square of the distance. Because [latex]\scriptsize {{F}_{\text{G}}}=W[/latex] and we know that [latex]\scriptsize W=mg[/latex], it follows that if [latex]\scriptsize W[/latex] increases, so will
*g*. - .

[latex]\scriptsize \begin{align*}F & =\text{G}\displaystyle \frac{{{{m}_{1}}{{m}_{2}}}}{{{{d}^{2}}}}\\ & =\left( {6.67\times {{{10}}^{{-11}}}} \right)\left( {\displaystyle \frac{{(6\times {{{10}}^{{24}}})(1\times {{{10}}^{{15}}})}}{{{{{(1.3\times {{{10}}^{{11}}})}}^{2}}}}} \right)\\ & =2.37\times {{10}^{7}}\text{ N}\end{align*}[/latex]

- .
- .

[latex]\scriptsize \begin{align*}{{F}_{\text{E}}} & =\text{G}\displaystyle \frac{{{{m}_{1}}{{m}_{2}}}}{{{{d}^{2}}}}\\ & =\left( {6.67\times {{{10}}^{{-11}}}} \right)\left( {\displaystyle \frac{{(70)(6\times {{{10}}^{{24}}})}}{{{{{(6.4\times {{{10}}^{6}})}}^{2}}}}} \right)\\ & =683.7\text{ N}\end{align*}[/latex] - His weight would be bigger on the new planet. The new planet has the same radius as the Earth, but a larger mass. Since the mass of the planet is proportional to the force of gravity (or the weight), the man’s weight must be larger.
- .

[latex]\scriptsize \begin{align*}{{F}_{\text{E}}} & =\text{G}\displaystyle \frac{{{{m}_{1}}4{{m}_{2}}}}{{{{d}^{2}}}}\\ & =4{{F}_{\text{E}}}\\ & =(683.7)(4)\\ & =2734.8\text{ N}\end{align*}[/latex]

- .
- Remember that mass is a measure of the amount of matter something consists of. The astronaut’s mass would therefore be the same on Earth and on Saturn, regardless of the size of Saturn relative to the Earth. So, we can calculate the astronaut’s mass using the gravitational acceleration as on Earth:

[latex]\scriptsize \begin{align*} & {{W}_{{\text{Earth}}}}=m{{g}_{{\text{Earth}}}}\\ & \therefore m =\displaystyle \frac{W}{g}\\ & =\displaystyle \frac{{750}}{{9.8}}\\ & =76.5\text{ kg}\end{align*}[/latex]

### Media Attributions

- QR_Code_PSL2SO22U3_1
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- Parts of the text in this unit were sourced from Siyavula Physical Science Gr 11 Learner’s Book, p. 109, released under a CC-BY licence at https://www.siyavula.com/read/science/grade-11/newtons-laws/02-newtons-laws-04. ↵

the acceleration caused by gravity acting on a free-falling object

a force between two objects because of their mass

an object that is acted on only by the force of gravity

a measure of how hard gravity is pulling on an object

the amount of matter in an object