Question

A satellite orbits a planet of unknown mass in a circle of radius $\mathbf{2}\mathbf{.}\mathbf{0}\mathbf{\times }{\mathbf{10}}^{\mathbf{7}}\mathbf{}\mathbf{m}$. The magnitude of the gravitational force on the satellite from the planet is $\mathbf{\text{F}}\mathbf{=}\mathbf{80}\mathbf{N}$.

(a) What is the kinetic energy of the satellite in this orbit?

(b) What would F be if the orbit radius were increased to$\mathbf{3.0}\mathbf{\times }{\mathbf{10}}^{\mathbf{7}}\mathbf{}\mathbf{m}$ ?

Short answer

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

Step-by-step solution

Listing the given quantities

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

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

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

Understanding the concept of gravitational force and kinetic energy

Using the given gravitational force and kinetic energy formula, we can findthekinetic 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:

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

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

(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),

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

(b) Calculation of Gravitational force when the orbit radius is increased 

$\begin{array}{c}{\text{F}}_{\text{1}}\propto \frac{\text{1}}{{\text{r}}_{\text{1}}^{\text{2}}}\\ {\text{F}}_2\propto \frac{\text{1}}{{\text{r}}_2^{\text{2}}}\\ \frac{{\text{F}}_{\text{1}}}{{\text{F}}_{\text{2}}}=\frac{{\text{r}}_{\text{2}}^{\text{2}}}{{\text{r}}_{\text{1}}^{\text{2}}}\\ \frac{{\text{F}}_{\text{1}}}{{\text{F}}_{\text{2}}}=\frac{{(2\times {10}^7\text{\hspace{0.17em}m})}^2}{{(3\times {10}^7\text{\hspace{0.17em}m})}^2}\\ \frac{{\text{F}}_{\text{1}}}{{\text{F}}_{\text{2}}}=\frac{4}{9}\end{array}$

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

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

Gravitational force when the orbit radius is increased to $3\times {10}^7\mathrm{m}\mathrm{is}{\mathrm{F}}_2=35.5\mathrm{N}$