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Expert-verified Found in: Page 383 ### Fundamentals Of Physics

Book edition 10th Edition
Author(s) David Halliday
Pages 1328 pages
ISBN 9781118230718 # A satellite orbits a planet of unknown mass in a circle of radius ${\mathbf{2}}{\mathbf{.}}{\mathbf{0}}{\mathbf{×}}{{\mathbf{10}}}^{{\mathbf{7}}}{\mathbf{}}{\mathbf{m}}$. The magnitude of the gravitational force on the satellite from the planet is ${\mathbf{\text{F}}}{\mathbf{=}}{\mathbf{80}}{\mathbf{\text{\hspace{0.17em}}}}{\mathbf{N}}$. (a) What is the kinetic energy of the satellite in this orbit? (b) What would F be if the orbit radius were increased to ${\mathbf{3.0}}{\mathbf{×}}{{\mathbf{10}}}^{{\mathbf{7}}}{\mathbf{}}{\mathbf{m}}$ ?

1. Kinetic energy of the satellite $\mathrm{K}=8.0×{10}^{8}\mathrm{J}$
2. Gravitational force when the orbit radius is increased to $3×{10}^{7}\mathrm{m}\text{\hspace{0.17em}is\hspace{0.17em}}\left({\text{F}}_{\text{2}}\right)=35.5\text{\hspace{0.17em}}\mathrm{N}$
See the step by step solution

## Step 1: Listing the given quantities

${\text{r}}_{\text{1}}=2×{10}^{7}\mathrm{m}$

${\text{F}}_{\text{1}}\text{}=80\text{\hspace{0.17em}}\mathrm{N}$

${\text{r}}_{\text{2}}\text{}=3×{10}^{7}\mathrm{m}$

## Step 2: Understanding the concept of gravitational force and kinetic energy

Using the given gravitational force and kinetic energy formula, we can find the kinetic energy of the satellite ( ${{\mathbf{\text{K}}}}_{{\mathbf{\text{s}}}}$ ). Using ${\mathbf{\text{F}}}{\mathbf{\propto }}\frac{\mathbf{\text{1}}}{{\mathbf{\text{r}}}^{\mathbf{\text{2}}}}{\mathbf{}}{\mathbf{,}}$ we can find gravitational force ${{\mathbf{F}}}_{{\mathbf{2}}}$ when the radius is increased to ${{\mathbf{\text{3×10}}}}^{{\mathbf{\text{7}}}}{\mathbf{\text{\hspace{0.17em}m}}}$

Formula:

$\text{K=}\frac{\text{GMm}}{\text{2}\left(\text{r}\right)}$

$\text{F=}\frac{\text{GMm}}{{\text{r}}^{\text{2}}}$

## Step 3: (a) Calculation of kinetic energy of satellite

$\text{K=}\frac{\text{GMm}}{\text{2}\left({\text{r}}_{\text{1}}\right)}$

${\text{F}}_{\text{1}}\text{=}\frac{\text{GMm}}{{\text{r}}_{\text{1}}^{\text{2}}}$

${\text{F}}_{\text{1}}{\text{×r}}_{\text{1}}=\frac{\text{GMm}}{{\text{r}}_{\text{1}}}$

Using this in kinetic energy equation (1),

$\begin{array}{c}\text{K}=\frac{\text{1}}{\text{2}}{\text{F}}_{\text{1}}×{\text{r}}_{\text{1}}\text{}\\ =\frac{1}{2}80\text{\hspace{0.17em}N}×2×{10}^{7}\text{\hspace{0.17em}m}\\ =8.0×{10}^{8}\text{\hspace{0.17em}}\mathrm{J}\end{array}$

## Step 4: (b) Calculation of Gravitational force when the orbit radius is increased

$\begin{array}{c}{\text{F}}_{\text{1}}\propto \frac{\text{1}}{{\text{r}}_{\text{1}}^{\text{2}}}\\ {\text{F}}_{2}\propto \frac{\text{1}}{{\text{r}}_{2}^{\text{2}}}\\ \frac{{\text{F}}_{\text{1}}}{{\text{F}}_{\text{2}}}=\frac{{\text{r}}_{\text{2}}^{\text{2}}}{{\text{r}}_{\text{1}}^{\text{2}}}\\ \frac{{\text{F}}_{\text{1}}}{{\text{F}}_{\text{2}}}=\frac{{\left(2×{10}^{7}\text{\hspace{0.17em}m}\right)}^{2}}{{\left(3×{10}^{7}\text{\hspace{0.17em}m}\right)}^{2}}\\ \frac{{\text{F}}_{\text{1}}}{{\text{F}}_{\text{2}}}=\frac{4}{9}\end{array}$

Using the given value of ${\text{F}}_{\text{1}}$

$\begin{array}{c}{\text{F}}_{\text{2}}=\frac{4×80\text{\hspace{0.17em}N}}{9}\\ =35.5\text{\hspace{0.17em}}\mathrm{N}\end{array}$

Gravitational force when the orbit radius is increased to $3×{10}^{7}\mathrm{m}\text{\hspace{0.17em}}\mathrm{is}\text{\hspace{0.17em}}{\mathrm{F}}_{2}=35.5\mathrm{N}$ ### Want to see more solutions like these? 