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...

\(\frac{\text{1}}{\text{9}}F\)

\(\frac{\text{1}}{\text{3}}F\)

\(\text{3}F\)

\(\text{9}F\)
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\(9F\)
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
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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*}
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*}
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.