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A spaceship is on a straight-line path between earth and moon. At what distance from the center of Earth is the net gravitational force on the ship is zero? (mass of Earth $M_E$, mass of moon $M_M$ and $d$ is the distance between center of Earth and the moon)

\[G\frac{M_Em}{r^2}=G\frac{M_Mm}{{\left(d-r\right)}^2}\to M_E{\left(d-r\right)}^2=M_Mr^2\]

Let $q\equiv \frac{M_M}{M_E}\ ,\ x=\frac{r}{d}$ then we have ${\left(1-x\right)}^2=qx^2\to \left(1-q\right)x^2-2x+1=0$

\[\Rightarrow x_{1,2}=\frac{2\pm \sqrt{4-4\sqrt{1-q}}}{2\left(1-q\right)}=\frac{1}{1\pm \sqrt{q}}\ \ or\ \ \ r_{1,2}=\frac{d}{1\pm \sqrt{\frac{M_M}{M_E}}}\]

So one of the roots is less than d and the other greater than d.

Turksat $2A$ is a communication satellite, which orbits on a geostationary orbit. The period of the satellites orbiting on the geostationary orbits is about $24$ hours. Therefore, from locations on the surface of the earth, Turksat 2A appears motionless. Calculate the radius of the geostationary orbit $R_G$. (Assume this orbit is a circular orbit. $M_E\sim \ 6\times {10}^{24}{\rm kg}$ and $G=6.6\times {10}^{-11}\ {\rm N.}\frac{{{\rm m}}^{{\rm 2}}}{{\rm k}{{\rm g}}^{{\rm 2}}}$) .

\[\Sigma F_r=ma_r\to \ F_G=ma_r\to \ \ G\frac{M_Em}{R^2}=m\frac{v^2}{R}\ \]

\[\Rightarrow R={\left(\frac{GM_E}{\omega}\right)}^{\frac{1}{3}}\]

Recall that in circular motion $v=R\omega=R\frac{2\pi}{T}$, where $T$ is the period of the motion.

For a satellite with period of $24$ hours, the angular velocity is

\[\omega=\frac{2\pi}{T}=\frac{2\pi}{24\times 60\times 60}\sim \ 7.27\times {10}^{-5}\ \frac{{\rm rad}}{{\rm s}}\]

$\Rightarrow R=42150\ {\rm km}$ from center of the Earth.

Since the radius of earth is about $6400\,{\rm km}$, then the Turksat is about $35750{\rm \ km}$ from the surface of the earth.

The gravitational acceleration at the surface of a planet is $22.5\, {\rm m/}{{\rm s}}^{{\rm 2}}$. Find the acceleration at a height above the surface equal to the planet's radius.

\[a_g=G\frac{M}{{\left(R+h\right)}^2}\]

Where $R$ and $M$ are the planet's radius and mass, respectively. $G$ is the universal gravitational constant. Hence, at the surface level

\[a_g=G\frac{M}{R+h}=G\frac{M}{R+0}=22.5\ \]

Now find the $a_g$ at a distance $R$ above the surface.

\[a^{'}_g=G\frac{M}{R+R}=\frac{1}{2}\left(G\frac{M}{R}\right)=\frac{1}{2}\left(22.5\right)=11.25\frac{{\rm m}}{{{\rm s}}^{{\rm 2}}}\]

The radius of a spherical planet is $R$ and its mass is $M$. At what distance above the planet's surface will the acceleration of gravity be exactly one tenth of its value at the surface?

\[g_r=\frac{1}{10}g_R\Rightarrow G\frac{M}{{\left(r+R\right)}^2}=\frac{1}{10}G\frac{M}{R^2}\]

\[\Rightarrow \ 10R^2={\left(r+R\right)}^2\]

Taking the square root of both sides, we obtain two solution.

\[\Rightarrow \sqrt{10R^2}=\sqrt{{\left(r+R\right)}^2}\Rightarrow \pm \sqrt{10}R=r+R\ \]

\[\Rightarrow \left\{ \begin{array}{rcl}

+\sqrt{10}R & = & r+R\Rightarrow r=R\left(\sqrt{10}-1\right)=2.16R \\

-\sqrt{10}R & = & r+R\Rightarrow r=-R(\sqrt{10}+1) \end{array}

\right.\]

The second result is not accepted since $r<0$!

You are on the home planet of the Klingon Empire. You drop a ball of mass $1.24\, {\rm kg}$ from a $334\, {\rm m}$ tower and it takes $6.54\, {\rm s}$ to reach the ground. The diameter of the Klingon home world is $1.55$ times the diameter of the earth. What is the mass of the Klingon home world?

\[y=\frac{1}{2}gt^2+v_0t+y_0\Rightarrow g=\frac{2y}{t^2}=\frac{2\times 334}{{\left(6.54\right)}^2}=15.61\frac{{\rm m}}{{{\rm s}}^{{\rm 2}}}\]

Where we have chosen the initial position of the ball as the base so $y_0=0$.

\[g=\frac{GM}{R^2}\ ,\ {\rm where\ R\ is\ the\ radius\ of\ the\ }{\rm planet}\]

\[M=\frac{gR^2}{G}=\frac{g{\left(1.55R_e\right)}^2}{G}=\frac{15.61\times {\left(1.55\times 6.37\times {10}^6\right)}^2}{6.67\times {10}^{-11}}=2.28\times {10}^{25}\ {\rm kg}\]

What is the speed of a satellite of mass $ m=1.0 \times 10^{4}\, {\rm kg}$ on a circular orbit around the moon at a distance of ${\rm 4}{{\rm R}}_{moon}$ from the moon's surface?

\[F_G=F_r\Rightarrow \frac{GM_Mm}{r^2}=\frac{mv^2}{r}\Rightarrow v=\sqrt{\frac{GM_M}{r}}=\sqrt{\frac{GM}{4R_M}}=\frac{1}{2}{\left(g_{moon}\right)}^{\frac{1}{2}}\]

Astronauts put their spaceship into orbit about a planet. They find that the acceleration of gravity at their orbital altitude is half that at the planet's surface. How far above the planet's surface are they orbiting? Answer in terms of the radius of the planet.

\[a_g=G\frac{M}{R^2}\]

And at the height of $h$ above its surface is

\[a_g=G\frac{M}{{\left(R+h\right)}^2}\]

In this problem, we know that $a_g\left(h\right)=\frac{1}{2}a_g(R)$ so

\[G\frac{M}{{\left(R+h\right)}^2}=\frac{1}{2}G\frac{M}{R^2}\Rightarrow {\left(R+h\right)}^2=2R^2\]

Taking square root from each side, we obtain

\[R+h=\pm \sqrt{2}R\Rightarrow \ \left\{ \begin{array}{lcl}

R+h=+\sqrt{2}R\Rightarrow h=\left(\sqrt{2}-1\right)R\ \ & , & \ \ {\rm I} \\

R+h=-\sqrt{2}R\Rightarrow h=-\left(1+\sqrt{2}\right)R\ \ & , & \ \ {\rm II} \end{array}

\right.\]

The answer ${\rm II}$ isn't correct since $h$ is negative. Therefore, the orbital altitude of the Astronauts is $h=0.41\ R$

Consider the ring-shaped body of radius $a$ and mass $M$ shown in the figure. A particle of mass $m$ is placed a distance $x$ from center of the ring, along a line through the center of the ring and perpendicular to it.

(a) Calculate the gravitational potential energy $U$ of this system. Take the potential energy to be zero when the two objects are very far apart. Recall that we can write $dU=-\frac{Gm}{r}dM$ where $r={\left(x^2+a^2\right)}^{\frac{1}{2}}$ as we integrate around the ring.

(b) Show that your answer in part (a) reduces to the point masses result when $x$ is much larger than the radius $a$ of the ring.

(c) Use $F_x=-dU/dx$ to find the magnitude and direction of the force on the particle.

(d) Show that your answer in part (c) reduces to the expected result when $x$ is much larger than $a$. What are the values of $U$ and $F_x$ when $x=0$ and why?

\[U=-G\frac{m_1m_2}{r^2}\]

First calculate the gravitational potential energy of an infinitesimal element of the ring $dM$ and mass $m$ then integrate it around the ring to find the total potential.

\[dU=-\frac{Gm}{r}dM\to U=-\frac{Gm}{r}\int{dM}=-\frac{GmM}{\sqrt{x^2+a^2}}\]

(b) If $x\gg a$, then using the binomial theorem ${\left(1+u\right)}^n=1+nu+\dots \ $ ($\left|u\right|$ must be smaller than $1$), we obtain

\[U=-\frac{GmM}{\sqrt{x^2+a^2}}=-\frac{GmM}{x{\left(1+\frac{a^2}{x^2}\right)}^{\frac{1}{2}}}=-\frac{GmM}{x}{\left(1+\frac{a^2}{x^2}\right)}^{\frac{1}{2}}=-\frac{GmM}{x}\left(1-\frac{a^2}{2x^2}\right)\]

(c) \[F_x=-\frac{dU}{dx}=GmM\left(-\frac{1}{2}\frac{2x}{{\left(x^2+a^2\right)}^{\frac{3}{2}}}\right)=GmM\frac{x}{{\left(x^2+a^2\right)}^{\frac{3}{2}}}\]

If we assume $x\gg a$ then

\[GmM\frac{x}{{\left(x^2+a^2\right)}^{\frac{3}{2}}}=\frac{GmM}{x^2}{\left(1+\frac{a^2}{x^2}\right)}^{-\frac{3}{2}}=\frac{GmM}{x^2}\left(1-\frac{3}{2}\frac{a^2}{x^2}+\dots \right)\cong \frac{GmM}{x^2}\]

(d) \[U\left(x=0\right)=-GmM/a\]

\[F_x\left(x=0\right)=GmM\left[\frac{x}{{\left(x^2+a^2\right)}^{\frac{3}{2}}}\right]=0\]

The contribution from $dM$ around the ring cancels the net force on the particle.

Category : Gravitation

MOST USEFUL FORMULA IN GRAVITY:

Newton's law of gravity:

\[F=-G\frac {m_1m_2}{r^2}\]

Universal gravitational constant:

\[G=6.67\times 10^{-11}\,\mathrm {N.m^{2}/kg^2}\]

Gravitational potential energy:

\[U(r)=-\frac{GmM}r\]

Number Of Questions : 8

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