# AP Physics 1: Circular Motion and Gravitation Practice Problems +Answers

Over 20 multiple-choice questions on circular motion and gravitation which appear on the AP Physics 1 exam are provided with detailed explanations.

# AP Physics: Circular Motion Solved Problems

Problem (1): A motorcycle weighing $200\,{\rm kg}$ turning around an unbanked circular track of radius $12.5\,{\rm m}$ at a constant speed of $56\,{\rm km/h}$. What is the magnitude of the net force applying on the motorcycle?

(a) $3.844\times 10^{3}\,\rm N$        (b) $5.0176\times 10^{4}\,\rm N$
(c) $248\rm N$                                  (d) $3.844\times 10^{4}\,\rm N$

Solution: Because the motorcycle moves around a circular path at a constant speed $v$, so it undergoes a centripetal force whose magnitude is given by the following formula $F_c=m\frac{v^2}{r}$ Plugging the numerical values into above, we get $F_c=200\times \frac{(15.5)^2}{12.5}=3844\,{\rm N}$ Keep in mind that usually in the AP physics 1 problems on circular motions, the constant speed is given in terms of $\rm \frac{km}{h}$. To use the centripetal acceleration or force formula above, we must convert it into the SI units of speed which is $\rm \frac ms$ by multiplying it by $\frac{10}{36}$. Thus, the given speed in SI units is $56\times \frac{10}{36}=15.5\,{\rm m/s}$ Thus, the correct answer is (a).

Problem (2): A $355-\rm g$ ball is fastened at the end of a $1.1-\rm m$ long rope and is whirled in a vertical circle at a constant speed of $3.75\,\rm m/s$. The tensions in the rope when the ball is at the top and the bottom of its path are, respectively, closest to

(a) $2\, ,\, 7$                (b) $8\, ,\, 1$
(c) $1\, ,\, 8$                (d) $1\, ,\, 9$

Solution: At the top and the bottom of the path the force are exerting on the ball is shown in the free-body diagram below. In both positions, the weight force is always down but at the top the tension is down and at the bottom is up. On the other hand, we know that the centripetal force is always directed toward the center of circle at each position. Applying Newton's second law, we get \begin{gather*} F_c=T+mg \quad \text{top} \\\\ \frac{mv^2}{r}=T+mg \\\\ \Rightarrow T=\frac{0.355\times (3.75)^2}{1.1}-(0.355)(9.8)\\\\ \Rightarrow \boxed{T=1.06\,\rm N} \\\\ F_c=T-mg \quad \text{Bottom}\\\\ \frac{mv^2}{r}=T-mg \\\\ \Rightarrow T=\frac{0.355\times (3.75)^2}{1.1}+(0.355)(9.8) \\\\ \Rightarrow \boxed{T=8.02\,\rm N}\end{gather*} Hence, the correct answer is (c).

Problem (3): A small bead of mass $20\,\rm g$ is attached to the end of light string $120\,\rm cm$-long and is deflected at an angle of $18\circ$. Once the bead is released, what is the speed of the bead in $\rm m/s$?

(a) $1.3$                (b) $3.2$
(c) $0.55$              (d) $3.5$

Solution: When the bead is held at your hand at an arbitrary angle, say $18\deg$, the tension in the string is $T\cos\theta=mg$, as you can see in the free-body diagram below. After releasing the bead, under the string's tension, moves around a semi-circle. As a consequence, it experiences a centripetal force whose magnitude is $T=F_c=\frac{mv^2}{\ell}$ where $\ell$ is the string's length. Substitute from the first equation the tension as $T=\frac{mg}{\cos\theta}$, into the second one, and solve for the speed $v$ \begin{align*} \frac{mg}{\cos\theta}&=\frac{mv^2}{\ell}\\\\ \Rightarrow v&=\sqrt{\frac{g\ell}{\cos\theta}}\\\\&=\sqrt{\frac{1.2\times 9.8}{\cos 18^\circ}}\\\\&=3.5\quad\rm m/s\end{align*} Thus, the correct answer is (d).

Problem (4): In a horizontal circle of radius $85\,{\rm cm}$, a constant force of $245\,\rm N$ is applied to a $2.5\,\rm kg$ object as it revolves around the track. The speed of the object (in $\rm m/s$) is closest to

(a) $91.3$            (b) $9.13$
(c) $83.3$             (d) $8330$

Solution: All information required to use the centripetal force formula $F_c=\frac{mv^2}{r}$ is given but the object's speed $v$. Thus, rearranging this for $v$ and substituting the known values into it, we get \begin{align*} v&=\sqrt{\frac{rF_c}{m}}\\\\ &= \sqrt{\frac{0.85\times 245}{2.5}}\\\\&=\boxed{9.13\quad \rm m/s}\end{align*} Thus, the correct choice is (b).

Problem (5): At the end of a very light string $2-\rm m$-long, a $500-\rm g$ toy is attached and whirled around a circle on a frictionless surface. Assume the maximum tension that the string can withstand is $350\,\rm N$ without breaking. In this situation, what is the maximum speed the toy can have in $\rm m/s$?

(a) $26$            (b) $37$
(c) $19$            (d) $42$

Solution: The toy whirled around a circular frictionless path so that it experiences an inward centripetal force which is provided by the tension in the string. The magnitude of this centripetal force which equals to the tension is obtained by $T=F_c=\frac{mv^2}{r}$. Substituting the numerical values into above formula, and solving for $v$, we get \begin{align*} v^2&=\frac{rT}{m}\\\\&=\frac{2\times 350}{0.5}\\\\&=\sqrt{1400}\\\\ \Rightarrow v&=37.5\quad\rm m/s\end{align*} which is closest to (b).

Problem (6): A race road is designed with a curve of radius $100\,\rm m$ and banked at an angle of $45^\circ$. What maximum speed a race car can have on this road without friction in $\rm m/s$?

(a) $25$            (b) $36$
(c) $31$            (c) $69$

Solution

Problem (7): A $1000-\rm kg$ motorcycle is moving at a constant speed of $30\,\rm m/s$ around a horizontal unbanked curve whose radius is $100\,\rm m$. To keep the car from sliding, how much friction force is required in $\rm N$?

(a) $9000$            (b) $8000$
(c) $7500$            (d) $6000$

Solution: The motorcycle wants to turn around a circular path, so a centripetal force toward the center of the circle is applied to it whose magnitude is obtained by $F_c=\frac{mv^2}{r}$. This force is provided by the friction force. Without friction, the motorcycle slides and deviates from the main track which is circular. Substituting the given numerical values into above, we get \begin{align*} F_c&=\frac{mv^2}{r}\\\\ &=\frac{1000\times 30^2}{100}\\\\&=9000\quad \rm N \end{align*} Thus, the correct answer is (a).

Problem (8): We want to create artificial gravity inside a large rotating hollow cylinder of radius $190\,\rm m$ by rotating it about its longitudinal axis. If we want the magnitude of this artificial gravity to be the same as the normal gravity, the time taken for each rotation of such a system is closest to?

(a) $32\,\rm s$            (b) $28\,\rm s$
(c) $30\,\rm s$            (d) $19\,\rm s$

Solution: Any object in a rotating system experiences a centripetal acceleration whose magnitude is found by $a_c=\frac{v^2}{r}$, where $v$ and $r$ are the tangential constant speed and radius of the circular system (path), respectively.

In this case, we want the magnitude of the artificial gravity to be the same as the Earth's gravitational acceleration that is $g=9.8\,\rm m/s^2$. Substituting this for $a_c$ and solving for $v$, we get the speed required to complete one full circle $g=\frac{v^2}{r} \Rightarrow v=\sqrt{gr}$

On the other hand, we know that each full rotation around a circle of radius $r$ corresponds a traveled distance of circumference of that circle, i.e., $\ell=2\pi r$. Hence, the time required to complete a full revolution, $T$, is obtained using the definition of average speed $v=\frac{\ell}{t}$ where $t$ is the time taken to cover the distance $\ell$. $T=\frac{\ell}{v}=\frac{2\pi r}{v}$ Substituting the constant velocity obtained above into this time taken formula, we get \begin{align*} T&=\frac{2\pi r}{\sqrt{rg}}\\\\ &=2\pi\sqrt{\frac{r}{g}}\\\\ &=2\pi\sqrt{\frac{190}{9.8}}\\\\ &=27.7\quad \rm s\end{align*} In the second line, to eliminate the radical from denominator, we rationalize the fraction by multiplying the numerator and denominator by $\sqrt{r}$. Thus, the correct choice is (b), since it is closest to the obtained value.

Problem (9): A car moving along a curved track, as shown in the figure below. At the lowest point of the curve, what is the direction of the normal force exerted by the road on the car?

(a) Upward            (b) Downward
(c) Forward            (d) Backward

Solution: When an object is in contact with a surface, a force is applied to it by the surface that is perpendicular. This force is known as normal force. At the lowest point of the track, the normal force is toward the center of the circular track. Thus, it is directed upward. Hence, the correct answer is (a).

Problem (10): A car is moving on a road with many hills and valleys. Assume the radius of these roughnesses to be $R$. At what point on the track, the driver most likely feel weightlessness?

(a) At the top of the hill when $v=\sqrt{Rg}$.
(b) At the bottom of the valley when $v>\sqrt{Rg}$
(c) At the top of the hill when $v<\sqrt{Rg}$
(d) At the bottom of the valley when $v=\sqrt{Rg}$

Solution: In physics, weightlessness is when the object loses contact with the surface. In this situation, the normal force becomes zero. At the hills, the normal force is up, and the weight force is downward. Taking up as positive and applying Newton's second law, the net force, which is equal to the centripetal force, at the top of hills is written as \begin{gather*} mg-F_N=\frac{mv^2}{R} \\\\\Rightarrow F_N=mg-\frac{mv^2}{R}\end{gather*} Similarly, at the bottom of the valley, the net force is \begin{gather*} F_N-mg=\frac{mv^2}{R} \\\\\Rightarrow F_N=mg+\frac{mv^2}{R}\end{gather*} Keep in mind that the centripetal force is always toward the center of the circle, for this reason at the hills is down and at the valleys is up.

As you can see, the normal force at the valleys can never be removed (it is always the sum of two nonzero forces) but at the hills it is possible. So, setting $F_N=0$ at the hills, we get $v=\sqrt{Rg}$ If the car's speed at the top of the hill is this amount, then the driver will feel weightlessness. Hence, the correct answer is (a).

Problem (11): A car is traveling along a curved path of radius $r$ at a constant speed $v$ as shown below. What is the apparent weight of the driver at the bottom of the curved path?

(a) $mg-mv^2/r$            (b) $mg$
(c) $mg+mv^2/r$            (d) none of these

Solution: the car moves along a circular path, so it undergoes a centripetal force toward the center of the circle. As shown in the free-body diagram below, two external forces are applied to the driver at this position; a downward weight force, and an upward normal force. Take up as positive and apply Newton's second law at this point \begin{gather*} F_N-mg=\frac{mv^2}{r} \\\\ \Rightarrow F_N=mg+\frac{mv^2}{r}\end{gather*} In all AP Physics 1 circular motion problems the apparent weight means the same as the normal force. Thus, the correct answer (c).

Problem (12): A bead is placed on a turntable that is rotating clockwise as in the figure below. Assume after a while, this turntable starts to decrease its spinning speed due to friction in its axis until it ceased. Which arrow best depicts the net force acting on the bead at the instant shown?

(a) $\downarrow$                (b) $\uparrow$
(c) $\searrow$                (d) $\nearrow$
(e) $\rightarrow$

Solution: The bead is rotating, so a centripetal force toward the center of the circle applies to it which at point $P$ is rightward. On the other side, the turntable is slowing down which means an opposing force (say, friction in the axis) must be applied to the bead. At point $P$, the bead's velocity is upward, so the opposing force must be downward to create a slowing down motion.

Consequently, the resultant of these two force vectors, which is obtained by the graphical method, is toward $\searrow$. Hence, the correct answer is (c).

## AP Physics: Gravitation Solved Problems

Problem (13): A satellite weighing $M$ takes a time of $T$ to complete a revolution around a planet. Now assume that the satellite had three times as much mass. What changes would occur for the time of circling around the planet at the same altitude?

(a) $2T$            (b) $4T$
(c) $T$                (d) $\frac{T}{2}$

Solution: One of the most common problems on circular motion and gravitation in the AP Physics 1 exam is about whirling a satellite around a planet. In such AP physics questions, the inward centripetal force that the satellite experiences is provided by the gravity force between the satellite and the planet. $\frac{GmM}{r^2}=\frac{mv^2}{r}$ where $m$ and $M$ refer to the masses of satellite and planet, respectively.

Usually, asked for the satellite's speed or period in the orbit. In this question, we consider the case of time orbiting around the planet.

The satellite travels a perimeter of a circle of radius $r$ at a constant speed $v$ in the orbit. This trip takes $T$ which is found using the definition of average speed as below $T=\frac{2\pi r}{v}$ Substituting this into the previous equation and rearranging it, we get $T^2=\frac{4\pi^2}{GM}r^3$ This is known as Kepler's third law. As you can see, the mass of the satellite does not appear in the time period formula. Hence, by changing the satellite's mass, its period around the planet does not change. So, the correct answer is (c).

Problem (14): An artificial satellite revolving around the Earth completes each orbit in $120$ minutes. At what altitude is this satellite located in $\rm km$?

(a) $8033$                (b) $9000$
(c) $6033$                (d) $10000$

Solution: The orbital period $T$ of a satellite (or a planet) at altitude $r$ circling around an object of mass $M$ is found using Kepler's third law as below formula $r^3=\frac{GM}{4\pi^2}T^2$ where with a good approximation we can set, in the AP Physics tests, $\frac{GM}{4\pi^2}=10^{13}$. In this question, the orbital period is given in minutes, which must be converted in seconds by multiplying it by $60$. $T=120\times 60=7200\,\rm s$ Plugging this into the previous formula and  taking the cube root of both sides, we get \begin{align*} r^3 &=\frac{GM}{4\pi^2}T^2\\\\&=10^{13}\times (7200)^2\\\\&=5.184\times 10^{20} \\\\ & \text{Take the cube root of both sides} \\\\ r&=\left(5.184\times 10^{20}\right)^{\frac 13} \\\\ &=8.033\times 10^6\quad\rm m\end{align*} Note that this result is in meters. By diving it by $1000$ it is converted to $\rm km$. Thus, the answer is $8033\,\rm km$ which is (a).

Problem (15): A satellite with a circular speed of $v$ revolves around a planet of mass $M$. Now assume if the planet had half as much as mass, how would the orbital speed of the satellite change at the same altitude (distance)?

(a) $\frac{v}{\sqrt{2}}$            (b) $\frac{v}{2}$
(c) $\sqrt{2}v$                                 (d) $2v$

Solution: The satellite moves around the planet in a circular orbit, so it experiences an inward centripetal force which is provided by the attraction gravity force between the Earth and the satellite. Without this gravity force, the satellite cannot whirl around the orbit. Thus, we have \begin{gather*} \frac{GmM_E}{r^2}=\frac{mv^2}{r}\\\\ \Rightarrow \quad v=\sqrt{\frac{GM}{r}} \end{gather*} where $m$ is the satellite's mass. As you can see, the satellite's speed in orbit is proportional to the square root of the planet's mass, i.e., $v\propto \sqrt{M}$. So, the ratio the of satellite's speed at two different cases mentioned above is \begin{gather*} \frac{v'}{v}=\sqrt{\frac{M'}{M}}=\sqrt{\frac{\frac 12 M}{M}}\\\\ \Rightarrow \quad v'=\frac{v}{\sqrt{2}} \end{gather*} where prime denotes the case when the planet's mass is half. Hence, if the planet had half as much as mass, then its speed in the orbit could be a factor of as much as $\frac{v}{\sqrt{2}}$. Therefore, the correct answer is (a).

Problem (16): Assume two planets, say $A$ and $B$, orbiting an unknown star. The planet $A$ is $10$ times farther from the star than is planet $B$. The ratio of their speed $\frac{v_A}{v_B}$ is closest to?

(a) $0.4$                (b) $0.2$
(c) $0.3$                (d) $0.5$

Solution: In the previous question, we realized that the speed of an object of mass $m$ orbiting at a distance of $r$ around a mass $M$ is given by $v=\sqrt{\frac{GM}{r}}$ This formula shows us that the speed in orbit inversely proportional to the square root of distance from the mass $M$. Thus, their ratio is found to be as below \begin{align*} \frac{v_A}{v_B}&=\sqrt{\frac{r_B}{r_A}}\\\\&=\sqrt{\frac{r_B}{10\,r_B}}\\\\&=\frac{1}{\sqrt{10}}\\\\ &=\boxed{0.31} \end{align*} Hence, the correct answer is (c).

Problem (17): Two bodies attract each other with a force of $F$ gravitationally. Now suppose the masses doubled and the separation between them gets tripled. After these changes, what will the force be?

(a) $4$            (b) $\frac{1}{36}$
(c) $1$            (d) $2$

Solution: According to Newton's universal gravitation law, $F=\frac{m_1 m_2}{r^2}$, the gravitation force between two masses of $m_1$ and $m_2$ is proportional to their product and inversely proportional to the square of distance between them. Thus, after changes have taken place, the force $F'$ is a factor of as much as the original force $F$ which is found as \begin{align*} \frac{F'}{F}&=\frac{m'_1 m'_2}{m_1 m_2}\times \frac{r^2}{r'^2} \\\\ &=\frac{(\frac 12 m_1)(\frac 12 m_2)}{m_1 m_2}\times \frac{r^2}{(3r)^2}\\\\&=\frac{1}{36}\end{align*} Thus, $F'=\frac{1}{36}F$ the correct answer is (b).

Problem (18): Consider two small spheres of masses $8\,\rm kg$ and $12\,\rm kg$ that are placed $2\,\rm m$ far away from each other. What is the magnitude of gravitational force between them in $\rm N$? (Take $G=\frac{20}{3}\times 10^{-11}\,\rm \frac{N.m^2}{kg^2}$).

(a) $1.6\times 10^{-8}$            (b) $3.2\times 10^{-8}$
(c) $1.6\times 10^{-9}$            (d) $3.2\times 10^{-9}$

Solution: Two objects of masses $m_1$ and $m_2$ at a center-to-center distance of $r$, attract each other with a gravitational force of magnitude $F=G\frac{m_1 m_2}{r^2}$ where $\rm G$ is the gravitational constant. Substituting the given values into this formula, we get \begin{align*} F&=G \frac{m_1 m_2}{r^2}\\\\&=\left(\frac{20}{3}\times 10^{-11}\right)\times \frac{8\times 12}{2^2}\\\\&=\boxed{1.6\times 10^{-9}\quad \rm N} \end{align*} Hence, the correct answer is (c).

Problem (19): Two small balls from $20\,\rm m$ from each other, attract one another with a gravitational force of $10^{-8}\rm N$. If one of the balls is $60\,\rm kg$, then what is the mass of the other ball? (Take $G=\frac{20}{3}\times 10^{-11}\,\rm \frac{N.m^2}{kg^2}$)

(a) $10^2$                (b) $10^3$
(c) $5$                        (d) $50$

Solution: Apply Newton's law of universal gravitation formula and solve for the unknown mass \begin{gather*} F=G \frac{m_1 m_2}{r^2}\\\\ 10^{-8}=\left(\frac{20}{3}\times 10^{-11}\right)\times \frac{60\times m_2}{20^2} \\\\ \Rightarrow \boxed{m_2=1000\quad\rm kg} \end{gather*} Hence, the correct answer is (b).

Problem (20): The masses of Earth and the Moon are about $5.98\times 10^{24}\rm kg$ and $7.36\times 10^{22}\,\rm kg$, respectively. The separation between them is also $3.84\times 10^8\,\rm m$. The order of magnitude of the gravitational force between those is closest to?

(a) $10^{22}$                (b) $10^{25}$
(c) $10^{28}$                (d) $10^{31}$

Solution: In this problem, the order of magnitude of the force is asked, so it is not necessary to get the full answer. First, determine the order of magnitude of each of the known values, then use the universal gravitation law.

Problem (21): Two small objects of masses $m$ and $2m$ at a distance of $r$ applying a gravitational force of $F$ on each other. How many $F$ times is the magnitude of the gravitational force between two objects of masses $3m$ and $4m$ at a distance of $2r$?

(a) $\frac 13$        (b) $3$
(c) $\frac 23$        (d) $\frac 32$

Solution: According to the universal gravitation law, $F=G\frac{m_1 m_2}{r^2}$, the attraction force between two objects is proportional to the product of their masses and inversely proportional to the square of distance between them. If we denote the varied masses and distance by prime, then we have \begin{align*} \frac{F'}{F}&=\frac{m'_1}{m_1} \times \frac{m'_2}{m_2} \times \left(\frac{r}{r'}\right)^2 \\\\&=\frac{3m}{m}\times \frac{4m}{2m}\times \left(\frac{r}{2r}\right)^2 \\\\&=3\times 2\times \frac{1}{4}\\\\&=\boxed{\frac 32} \end{align*} Hence, the correct answer is (d).

Problem (22): The plot of the gravitational force between two small objects in terms of distance between them is depicted below. What is $r$ in this plot?

(a) $8$                    (b) $12$
(c) $16$                (d) $24$

Solution: From the plot, we can find the force between those two masses at an arbitrary distance.

Problem (23): Two similar particles of masses $m$ at a distance of $r$ exert a gravitational force of magnitude $F$ on each other. What percent of the mass of one should be removed and added to the other so that at the previous distance, the magnitude of the gravitational force between them reduces $25\%$?

(a) $75$            (b) $25$
(c) $50$            (d) $33$

Solution: Assume $x$ as much is removed from one and is added to the other. After this manipulation, the masses becomes $m'_1=m-x$ and $m'_2=m+x$. On the other hand, the force between them $F'$, as a result, gets reduced $25\%$ as much as the original force $F$. This means $F'=F-0.25F=0.75F$ Now, form their ratio and apply universal gravitation law to find the unknown $x$ \begin{gather*} \frac{F'}{F}=\frac{m'_1}{m_1}\times \frac{m'_2}{m_2} \\\\ \frac 34=\frac{m-x}{m}\times \frac{m+x}{m}\\\\ \frac{3}{4}=\frac{m^2-x^2}{m^2}\\\\ 3m^2=4m^2-4x^2 \\\\ \Rightarrow 4x^2=m^2 \\\\ \Rightarrow x=\frac{m}{2} \end{gather*} This computation shows us that we must subtract by $50\%$ from one and add to the other. Thus, the correct answer is (c).

Problem (24): Imagine a point between the Moon and Earth where if an object is placed there, the net force applied on it from the Moon and Earth be zero. How many times is the distance of that point from the Earth's center relative to the Moon's center? (Take Earth's mass to be $81$ times the Moon's mass).

(a) $9$            (b) $10$
(c) $80$          (d) $81$

Solution: Assume the point lies at a distance of $r_E$ from the Earth and $r_M$ from the Moon. We want at this point the gravitational force on an imaginary mass $m$ is zero. So, the gravitational forces $F_E$ from the Earth and $F_M$ from the Moon on $m$ must be balanced.

\begin{gather*} F_E=F_M \\\\ \frac{GmM_E}{r_E^2}=\frac{GmM_M}{r_M^2} \\\\ \Rightarrow \frac{M_E}{r_E^2}=\frac{M_M}{r_M^2}\\\\ \rm{set \ M_E=81M_M} \\\\ \Rightarrow \frac{81M_M}{r_E^2}=\frac{M_M}{r_M^2} \\\\ \Rightarrow r_E^2=81\times r_M^2 \\\\ \Rightarrow \boxed{r_E=9r_M}\end{gather*} Hence, the correct answer is (a).

Problem (25): An spacecraft of mass $3\times 10^4\,\rm kg$ is in between the distance of Earth and the Moon. What is the net gravitational force applying on this spacecraft at this location? (Take the Earth's mass, $6\times 10^{24}\,\rm kg$, the Moon's mass $7.5\times 10^{22}\,\rm kg$, the distance between Earth and the Moon $4\times 10^5\,\rm km$, and $G=\frac{20}{3}\times 10^{-11}\,\rm \frac{N.m^2}{kg^2}$)

(a) $303.75$            (b) $296.25$
(c) $607.5$                (d) $592.5$

Solution: The spacecraft is positioned halfway between the Earth and the Moon. Thus, its distance from both is $d=\frac{4\times 10^5}{2}=2\times 10^5\rm km$ Now, using the universal gravitational law find the force exerted by the Earth and the Moon on the spacecraft.

\begin{align*} F_E&=G\frac{M_E m}{d^2}\\\\ &=\left(\frac{20}{3}\times 10^{-11}\right)\times \frac{(3\times 10^4)(6\times 10^{24})}{4\times 10^{16}}\\\\&=3\times 10^2 =300\,{\rm N} \\\\ F_M&=G\frac{M_M m}{d^2}\\\\ &=\left(\frac{20}{3}\times 10^{-11}\right)\times \frac{(3\times 10^4)(7.5\times 10^{22})}{4\times 10^{16}}\\\\&=3.75\,{\rm N} \end{align*} These two forces on the spacecraft are in the same line but in opposite direction, so the net force applying on the spacecraft is their subtraction. \begin{align*} F_{net}&=F_E-F_M \\\\ &=300-3.75\\\\&=\boxed{296.25\quad {\rm N}}\end{align*} Thus, the correct answer is (b).

Problem (26): The ESA wants to design a satellite to orbit around Earth $5$ times a day. What would be the radius of its orbit in $\rm m$, neglecting the presence of the Moon? (Assume $G=6.67\times 10^{-11}\,\rm \frac{N.m^2}{kg^2}$, and $M_{earth}=5.97\times 10^{24}\,\rm kg$)

(a) $3.45\times 10^7$            (b) $1.44\times 10^7$
(c) $2.11\times 10^7$            (d) $0.69\times 10^7$

Solution: '' $5$ times a day'' means $5$ full revolution every $24$ hours. Thus, one full revolution takes $\frac{24}{5}$ hours or $\frac{24}{5}\times 3600$ in seconds. This is the time period of this designed satellite. Now, apply Kepler's third law of motion to find the radius of orbit \begin{align*} r^3 &=\frac{GM}{4\pi^2}T^2\\\\&=10^{13}\times \left(\frac{24}{5}\times 3600\right)^2 \\\\ & \text{Take the cube root of both sides} \\\\ r&=1.44\times 10^7\quad \rm m \end{align*} Keep in mind that in all the AP physics gravitation questions, we can set $\rm \frac{GM}{4\pi^2}=10^{13}$ with a good approximation. Thus, the correct answer is (b).

Note: the formula of cube root is $a=\sqrt[3]{b} \ \text{or} \ a= b^{\frac 13}$

Author: Dr. Ali Nemati
Published: 1/1/2022