End of chapter exercises part 2

Gravitation

Textbook Exercise 2.11

An object attracts another with a gravitational force F. If the distance between the centres of the two objects is now decreased to a third ($\frac{\text{1}}{\text{3}}$) of the original distance, the force of attraction that the one object would exert on the other would become...

$\frac{\text{1}}{\text{9}}F$
$\frac{\text{1}}{\text{3}}F$
$\text{3}F$
$\text{9}F$

[SC 2003/11]

$9F$

An object is dropped from a height of $\text{1}$ $\text{km}$ above the Earth. If air resistance is ignored, the acceleration of the object is dependent on the...

mass of the object
radius of the earth
mass of the earth
weight of the object

[SC 2003/11]

weight of the object

A man has a mass of $\text{70}$ $\text{kg}$ 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 ( ${M}_{E}= \text{6} \times \text{10}^{\text{24}}\text{ kg}$, ${r}_{E}=\text{6} \times \text{10}^{\text{6}}\text{ m}$ )

Calculate the force between the man and the Earth.

\begin{align*} F_{E} & = \frac{Gm_{1}m_{2}}{d^{2}} \\ & = \frac{(\text{6,67} \times \text{10}^{-\text{11}})(\text{70})(\text{6} \times \text{10}^{\text{24}})}{(\text{6} \times \text{10}^{\text{6}})^{2}} \\ & = \text{778,2}\text{ N} \end{align*}

What is the man's weight on the new planet?

The mass of the new planet is four times the mass of the Earth so we get:

\begin{align*} F_{\text{planet}} & = \frac{Gm_{1}4m_{2}}{d^{2}} \\ & = 4F_{E} \\ & = (\text{778,2})(\text{4}) \\ & = \text{3 112,8}\text{ N} \end{align*}

Would his weight be bigger or smaller on the new planet? Explain how you arrived at your answer.

His weight is 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.

Calculate the distance between two objects, $\text{5 000}$ $\text{kg}$ and $\text{6} \times \text{10}^{\text{12}}$ $\text{kg}$ respectively, if the magnitude of the force between them is $\text{3} \times \text{10}^{\text{8}}$ $\text{N}$

\begin{align*} F & = \frac{Gm_{1}m_{2}}{d^{2}} \\ \text{3} \times \text{10}^{\text{8}} & = \frac{(\text{6,67} \times \text{10}^{-\text{11}})(\text{5 000})(\text{6} \times \text{10}^{\text{12}})}{d^{2}} \\ \text{3} \times \text{10}^{\text{8}}d^{2} & = \text{2,001} \times \text{10}^{\text{6}} \\ d^{2} & = \text{6,67} \times \text{10}^{-\text{3}} \\ & = \text{8,2} \times \text{10}^{-\text{2}}\text{ m} \end{align*}

The magnitude of the gravitational force which the astronaut experiences in the satellite.

We first work out the mass of the astronaut:

\begin{align*} F & = \frac{Gm_{1}m_{2}}{d^{2}} \\ \text{700} & = \frac{(\text{6,67} \times \text{10}^{-\text{11}})(\text{6} \times \text{10}^{\text{24}})m_{A}}{(\text{6 400} \times \text{10}^{\text{3}})^{2}} \\ \text{9,77}m_{A} & = \text{700} \\ m_{A} & = \text{71,6}\text{ kg} \end{align*}

Now we can work out the gravitational force in the satellite:

\begin{align*} F_{S} & = \frac{Gm_{1}m_{2}}{d^{2}} \\ & = \frac{(\text{6,67} \times \text{10}^{-\text{11}})(\text{6} \times \text{10}^{\text{24}})(\text{71,6})}{(\text{6 400} \times \text{10}^{\text{3}} + \text{1 600} \times \text{10}^{\text{3}})^{2}} \\ & = \text{448}\text{ N} \end{align*}

The magnitude of the gravitational force on an object in the satellite which weighs $\text{300}$ $\text{N}$ on Earth.

We first work out the mass of the object:

\begin{align*} F & = \frac{Gm_{1}m_{2}}{d^{2}} \\ \text{300} & = \frac{(\text{6,67} \times \text{10}^{-\text{11}})(\text{6} \times \text{10}^{\text{24}})m_{O}}{(\text{6 400} \times \text{10}^{\text{3}})^{2}} \\ \text{9,77}m_{O} & = \text{300} \\ m_{O} & = \text{30,7}\text{ kg} \end{align*}

Now we can work out the gravitational force in the satellite:

\begin{align*} F_{S} & = \frac{Gm_{1}m_{2}}{d^{2}} \\ & = \frac{(\text{6,67} \times \text{10}^{-\text{11}})(\text{6} \times \text{10}^{\text{24}})(\text{30,7})}{(\text{6 400} \times \text{10}^{\text{3}} + \text{1 600} \times \text{10}^{\text{3}})^{2}} \\ & = \text{192}\text{ N} \end{align*}

An astronaut of mass $\text{70}$ $\text{kg}$ on Earth lands on a planet which has half the Earth's radius and twice its mass. Calculate the magnitude of the force of gravity which is exerted on him on the planet.

The gravitational force on Earth is:

\[F_{E} = \frac{Gm_{E}m_{A}}{r_{E}^{2}}\]

On the planet we have twice the Earth's mass and half the Earth's radius:

\begin{align*} F_{P} & = \frac{Gm_{P}m_{A}}{r_{P}^{2}} \\ & = \frac{2Gm_{E}m_{A}}{\frac{r_{E}^{2}}{4}} \\ & = 8F_{E} \\ & = 8(70)(\text{9,8}) \\ & = \text{5 488}\text{ N} \end{align*}

Calculate the magnitude of the gravitational force of attraction between two spheres of lead with a mass of $\text{10}$ $\text{kg}$ and $\text{6}$ $\text{kg}$ respectively if they are placed $\text{50}$ $\text{mm}$ apart.

\begin{align*} F & = \frac{Gm_{1}m_{2}}{d^{2}} \\ & = \frac{(\text{6,67} \times \text{10}^{-\text{11}})(\text{10})(\text{6})}{(\text{50} \times \text{10}^{-\text{3}})^{2}} \\ & = \text{1,6} \times \text{10}^{\text{6}}\text{ N} \end{align*}

The gravitational force between two objects is $\text{1 200}$ $\text{N}$. What is the gravitational force between the objects if the mass of each is doubled and the distance between them halved?

The gravitational force is:

\[F_{1} = \frac{Gm_{1}m_{2}}{d^{2}}\]

If we double each mass and halve the distance between them we now have:

\begin{align*} F_{2} & = \frac{G(2m_{1})(2m_{2})}{(\text{0,5}d)^{2}} \\ & = \frac{4Gm_{1}m_{2}}{\text{0,25}d^{2}} \\ & = 16F_{1} \end{align*}

So the force will be $\text{16}$ times as much.

Calculate the gravitational force between the Sun with a mass of $\text{2} \times \text{10}^{\text{30}}$ $\text{kg}$ and the Earth with a mass of $\text{6} \times \text{10}^{\text{24}}$ $\text{kg}$ if the distance between them is $\text{1,4} \times \text{10}^{\text{8}}$ $\text{km}$.

\begin{align*} F & = \frac{Gm_{1}m_{2}}{d^{2}} \\ & = \frac{(\text{6,67} \times \text{10}^{-\text{11}})(\text{2} \times \text{10}^{\text{30}})(\text{6} \times \text{10}^{\text{24}})}{(\text{1,4} \times \text{10}^{\text{11}})^{2}} \\ & = \text{4,1} \times \text{10}^{\text{22}}\text{ N} \end{align*}

the mass of each object is doubled.

The gravitational force will be four times as much.

the distance between the centres of the objects is doubled.

The gravitational force will be one fourth as much or four times smaller.

the mass of one object is halved, and the distance between the centres of the objects is halved.

The gravitational force will be twice as much.

The gravitational acceleration $g$ is constant.

Agree

The weight of an object is independent of its mass.

Disagree. Weight is related to mass via the gravitational acceleration

G is dependent on the mass of the object that is being accelerated.

Disagree. G is a universal constant

What will his weight be on the surface of Saturn, which has a mass $\text{100}$ times greater than the Earth, and a radius $\text{5}$ times greater than the Earth?

On Earth we have:

\[a_{E} = G\frac{M_{E}}{R_{E}^{2}}\]

On Saturn we note that $M_{S} = 100M_{E}$ and $R_{S} = 5R_{E}$. So the gravitational acceleration on Saturn is:

\begin{align*} a_{S} & = G\frac{M_{S}}{R_{S}^{2}}\\ & = G\frac{100M_{E}}{25R_{E}^{2}} \\ & = 4G\frac{M_{E}}{R_{E}^{2}} \\ \therefore a_{S} &= 4a_{E} \end{align*}

So the weight of the astronaut on Saturn is:

\begin{align*} a_{S} & = 4a_{E} \\ & = 4(750) \\ & = \text{3 000}\text{ N} \end{align*}

What is his mass on Saturn?

On Earth his mass is:

\begin{align*} a_{E} & = mg_{E} \\ m & = \frac{a_{E}}{g_{E}} \\ & = \frac{\text{750}}{\text{9,8}} \\ & = \text{76,53}\text{ kg} \end{align*}

His mass on Saturn is the same as his mass on Earth, it is only his weight that is different.

Your mass is $\text{60}$ $\text{kg}$ in Paris at ground level. How much less would you weigh after taking a lift to the top of the Eiffel Tower, which is $\text{405}$ $\text{m}$ high? Assume the Earth's mass is $\text{6,0} \times \text{10}^{\text{24}}$ $\text{kg}$ and the Earth's radius is $\text{6 400}$ $\text{km}$.

We start with your weight on the surface of the Earth. The gravitational acceleration at the surface of the Earth is $\text{9,8}$ $\text{m·s$^{-2}$}$ and so your weight is:

\begin{align*} F_{g} & = mg \\ & = (\text{60})(\text{9,8}) \\ & = \text{588}\text{ N} \end{align*}

At the top of the Eiffel Tower the gravitational acceleration is:

\begin{align*} a_{o} & = G\frac{M_{\text{Earth}}}{d^{2}}\\ & = \frac{(\text{6,67} \times \text{10}^{-\text{11}})(\text{6,0} \times \text{10}^{\text{24}})}{(\text{6 400} \times \text{10}^{\text{3}} + \text{405})^{2}} \\ & = \text{9,77}\text{ m·s$^{-2}$} \end{align*}

Your weight is:

\begin{align*} F_{g} & = mg \\ & = (\text{60})(\text{9,77}) \\ & = \text{586,2}\text{ N} \end{align*}

So you would weigh $\text{1,8}$ $\text{N}$ less.

State Newton's law of universal gravitation.

Every body in the universe exerts a force on every other body. The force is directly proportional to the product of the masses of the bodies and inversely proportional to the square of the distance between them.

Use Newton's law of universal gravitation to determine the magnitude of the acceleration due to gravity on the Moon.

The mass of the Moon is $\text{7,4} \times \text{10}^{\text{22}}$ $\text{kg}$.

The radius of the Moon is $\text{1,74} \times \text{10}^{\text{6}}$ $\text{m}$.

\begin{align*} g & = \frac{F}{m_o} \\ & = \frac{G \frac{m_{\text{moon}}m_o}{r^2}}{m_2} \\ & = G\frac{m_{\text{moon}}}{r^2} \\ & = \frac{(\text{6,67} \times \text{10}^{-\text{11}})(\text{7,4} \times \text{10}^{\text{22}})}{(\text{1,74} \times \text{10}^{\text{6}})^2} \\ & = \text{1,63} \end{align*}

Will an astronaut, kitted out in his space suit, jump higher on the Moon or on the Earth? Give a reason for your answer.

He will be able to jump higher on the moon. The acceleration due to gravity is lower on the moon than on the Earth and so there is less gravitational force pulling him down on the moon.

End of chapter exercises part 1

Table of Contents

3.1 Chemical bonds

Maths

Science

Gravitation