# AP Physics C Mechanics: Unit 7 Practice Test — Gravitation

Our AP Physics C Mechanics unit 7 practice test covers gravitation. To answer these questions you should be familiar with Newton’s laws of Gravitation and their applications to different systems. Also, the motion of satellites, variation of gravitational acceleration with height, gravitational potential energy, and Kepler’s laws are included topics.

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 Question 1

### Find the force of gravity between two-point particles of mass $2.5 \ kg$ and $3.0 \ kg$ which are at a distance of $10 \ cm$ from each other.

 A $50.03×10^{-9} \ N$ B $50.03×10^{-11} \ N$ C $0.50×10^{-8} \ N$ D $5.03×10^{-9} \ N$ E $0.05×10^{-10} \ N$
Question 1 Explanation:
$F_g=\dfrac{Gm_1 m_2}{r^2}$

Using $G=6.67×10^{-11}, \ m_1=2.5 \ kg,m_2=3.0 \ kg$ and $r=0.1 \ m$ in the above equation gives,

$F_g=\dfrac{6.67×10^{-11}×2.5×3}{0.1^2} = 50.03×10^{-9} \ N$
 Question 2

### (The mass of each star is equal to $6 × 10^{28} \ kg$ and the $XY$ axis is in standard orientation)

 A $(9.68 \skew{2.3}\hat{i} - 29.20 \skew{4.5}\hat{j} )×10^{24} \ N$ B $(9.60 \skew{2.3}\hat{i} + 29.25 \skew{4.5}\hat{j} )×10^{25} \ N$ C $(29.20\skew{2.3}\hat{i} + 9.60 \skew{4.5}\hat{j} )×10^{27} \ N$ D $(6.78 \skew{2.3}\hat{i} - 26.38 \skew{4.5}\hat{j} )×10^{27} \ N$ E $(19.45 \skew{2.3}\hat{i} - 9.20 \skew{4.5}\hat{j} )×10^{26} \ N$
Question 2 Explanation:
Force on $J$ due to $K =\dfrac{Gm^2}{r^2}$

$=\dfrac{6.67×10^{-11}×36×10^{56}}{12.25×10^{18} }$

$=19.6×10^{27} \ N$ (along $-\skew{5}\hat{j}$)

Force on $J$ due to $L=\dfrac{Gm^2}{r^2}$

$=\dfrac{6.67×10^{-11}×36×10^{56}}{25×10^{18}}$

$=9.60×10^{27} \ N$ (along $\left(\dfrac{\skew{2}\hat{i}}{\sqrt{2}}-\dfrac{\skew{5}\hat{j}}{\sqrt{2}}\right)$)

Net force $=(6.78\skew{2}\hat{i}-26.38\skew{5}\hat{j} )×10^{27} \ N$
 Question 3

### $(M_e=6×10^{24} \ kg, \ R_e=6400 \ km)$

 A $2.05 \ hr$ B $1.52 \ hr$ C $1.75 \ hr$ D $1.88 \ hr$ E $2.15 \ hr$
Question 3 Explanation:
For a satellite moving in a circular orbit,

$\dfrac{m_s v^2}{r} = \dfrac{GM_e m_s}{r^2}$

$v_s=\sqrt{\dfrac{GM_e}{r}}$

Here $r=R_e+1,000 \ km$

Using $r=7.4×10^6 m$ and $M_e=6×10^{24} \ kg$

$v_s=\sqrt{\dfrac{6.67×10^{-11}×6×10^{24}}{7.4×10^6 }}=7,348 \ m/s$

Using $T=\dfrac{2πr}{v_s}$   gives   $T=6327.6 \ s=1.75 hr$
 Question 4

### Which of the below options shows the ratio obtained by Alex?

 A $0.54 ∶ 1$ B $0.49 ∶ 1$ C $0.45 ∶ 1$ D $0.60 ∶ 1$ E $0.36 ∶ 1$
Question 4 Explanation:
If $h$= Distance from the surface of the Earth then,

$g(h)=\dfrac{GM_e}{(R_e+h)^2}$

For point $A, h=2.5R_e-R_e=1.5R_e$

$g(h)=\dfrac{GM_e}{(R_e+1.5R_e )^2} =\dfrac{GM_e}{6.25R_e^2}$

For point $B, h=1.5R_e-R_e=0.5R_e$

$g(h)=\dfrac{GM_e}{(R_e+0.5R_e )^2} =\dfrac{GM_e}{2.25R_e^2}$

Ratio of acceleration $=\dfrac{GM_e}{6.25R_e^2 }×\dfrac{2.25R_e^2}{GM_e}=0.36 ∶ 1$
 Question 5

### A satellite is launched from the surface of a planet such that it escapes the gravitational field and travels with a speed of $10 \ km/s$ thereafter. Determine the velocity with which the satellite must be launched. (Take the mass of the planet as $3×10^{26} \ kg$ and the radius as $8×10^7 \ m$.)

 A $22.8 \ km/s$ B $24.4 \ km/s$ C $26.9 \ km/s$ D $29.5 \ km/s$ E $21.6 \ km/s$
Question 5 Explanation:
Total energy of the satellite far away from the planet $=\dfrac{1}{2} mv_{away}^2$

Total energy of the satellite on the surface of the planet $=\dfrac{1}{2} mv_{esc}^2 - \dfrac{GM_p m}{R_p}$

Equating the two equations gives,

$\dfrac{1}{2} mv_{away}^2=\dfrac{1}{2} mv_{esc}^2 - \dfrac{GM_p m}{R_p}$

$v_{away}^2=v_{esc}^2-\dfrac{2GM_p}{R_p}$

Using $v_{away}=10^4 \ m/s, M_p=3×10^{26} \ kg$ and $R_p=8×10^7 m$ gives,

$v_{esc}^2=v_{away}^2+\dfrac{2GM_p}{R_p}$

$v_{esc}^2=(10^4 )^2+\dfrac{2×6.67×10^{-11}×3×10^{26}}{8×10^7}$

$v_{esc}^2=10^8+5.00×10^8$

$v_{esc}^2=6.00×10^8$

$v_{esc}=2.44×10^4 \ m/s$

$v_{esc}=24.4 \ km/s$
 Question 6

### In which position does the planet travel with the fastest speed around the Sun?

 A $D$ B $C$ C $B$ D $A$ E None of the above
Question 6 Explanation:
Since, the Sun-Planet system is isolated the total mechanical energy is conserved

And total energy at any point on the trajectory is given by the expression,

$TE=\dfrac{1}{2} mv^2 - \dfrac{GMm}{r}$

For the speed (or kinetic energy) to be maximum, the term $\dfrac{GMm}{r}$ should be as large as possible (so that the difference remains a constant)

$\dfrac{GMm}{r}$ is maximum when $r$ is small

Thus, the planet has the highest speed at point $C$
 Question 7

### (Assume the mass of the planet to be 5×10^25 kg and the mass of the satellite to be 1.2 × 10^2 \kg)

 A $-16.24 × 10^6 \ J$ B $-10.17 × 10^6 \ J$ C $-11.02 × 10^6 \ J$ D $-15.82 × 10^6 \ J$ E $-13.34 × 10^6 \ J$
Question 7 Explanation:
For a satellite moving about a planet in a circular orbit,

$v=\sqrt{\dfrac{GM_p}{r}}→KE=\dfrac{1}{2} \dfrac{ GM_p m_s}{r}$

$PE=-\dfrac{GM_p m_s}{r}$

$TE=PE+KE$

$TE=-\dfrac{1}{2} \dfrac{GM_p m_s}{r}$

Using $M_p=5×10^{25} \ kg, r=1.5×10^{10} \ m, m_s=1.2×10^2 \ kg$

$TE=-\dfrac{1}{2} \dfrac{6.67×10^{-11}×5×10^{25}×1.2×10^2}{1.5×10^{10}}$

$TE=-13.34×10^6 \ J$
 Question 8

### The time period of two satellites, $LI$ and $MBII$, moving around the Earth is $24 \ hours$ and $2.5 \ hours$, respectively. The distance of the satellite $LI$ is approximately $36,000 \ km$ from the center of the Earth. What is the distance between $MBII$ satellite from the center of the Earth?

 A $6,324 \ km$ B $7,204 \ km$ C $8,122 \ km$ D $7,964 \ km$ E $7,684 \ km$
Question 8 Explanation:
According to Kepler’s 3rd law,

$T^2∝r^3$

Thus,

$\left(\dfrac{T_{MBII}}{T_{LI}} \right)^2=\left( \dfrac{r_{MBII}}{r_{LI}} \right)^3$

$\left(\dfrac{2.5}{24}\right)^2= \left(\dfrac{r_{MBII}}{36,000}\right)^3$

$\left( \dfrac{r_{MBII}}{36,000}\right)^3=\dfrac{1}{92.16}$

Solving for $r_{MBII}$ gives,

$r_{MBII}=\dfrac{36,000}{4.52}=7,964 \ km$
 Question 9

### (Hint: You might find the relation $\dfrac{dv}{dt} = v \dfrac{dv}{dx}$ helpful)

 A $v(x)=2 \sqrt{ \dfrac{GM}{R_e+h-x} - \dfrac{GM}{R_e+h}}$ B $v(x)=2 \sqrt{ \dfrac{GM}{R_e+h+x}}$ C $v(x)=2 \sqrt{ \dfrac{GM}{R_e+h-x}}$ D $v(x)=2 \sqrt{ \dfrac{GM}{R_e+h} - \dfrac{GM}{R_e+h+x}}$ E $v(x)=2 \sqrt{ \dfrac{GM}{R_e+h} + \dfrac{GM}{R_e+h+x}}$
Question 9 Explanation:
$\dfrac{dv}{dt} = \dfrac{GM}{r^2}$

Using the differential equation from the question gives,

$v \dfrac{dv}{dx} = \dfrac{GM}{r^2}$

Note: $r=R_e+(h-x)$

Solving the differential equation gives,

$\dfrac{v^2}{2}=C + \dfrac{GM}{R_e+h-x}$

At $x=0, \ v=0$ which implies $C= - \dfrac{GM}{R_e+h}$

$v(x)=2 \sqrt{ \dfrac{GM}{R_e+h-x} - \dfrac{GM}{R_e+h}}$
 Question 10

### What should be the speed v of the rover with which it should be launched from planet $A$ so that it reaches planet $B$ with a speed of $u = \sqrt{ \dfrac{4Gm}{R}}$?(Note: The values inside the circles shows the mass and the radius of the planet respectively.)

 A $\sqrt{\dfrac{1.66Gm}{R}}$ B $\sqrt{\dfrac{2.41Gm}{R}}$ C $\sqrt{\dfrac{3.38Gm}{R}}$ D $\sqrt{\dfrac{2.02Gm}{R}}$ E $\sqrt{\dfrac{4.56Gm}{R}}$
Question 10 Explanation:
Total energy of rover on planet

$A=\dfrac{1}{2} m_r v^2 - \dfrac{Gmm_r}{R} - \dfrac{2Gmm_r}{51.5R}$

$=\dfrac{1}{2} m_r v^2-1.04 \dfrac{Gmm_r}{R}$

Total energy of rover on planet

$B=\dfrac{1}{2} m_r \left(\dfrac{4Gm}{R}\right)-\dfrac{Gmm_r}{51R} - \dfrac{2Gmm_r}{1.5R}$

$=2 \dfrac{Gmm_r}{R-1.35} \dfrac{Gmm_r}{R}$

$=0.65 \dfrac{Gmm_r}{R}$

Since the total mechanical energy of the rover is a constant

$\dfrac{1}{2} m_r v^2-1.04 \dfrac{Gmm_r}{R}=0.65 \dfrac{Gmm_r}{R}$

$\dfrac{1}{2} m_r v^2=1.69 \dfrac{Gmm_r}{R}$

$v^2=\dfrac{3.38Gm}{R}$

$v=\sqrt{ \dfrac{3.38Gm}{R}}$
 Question 11

### Which of the below statements is TRUE regarding Kepler’s 2nd law?

 A The area swept per unit time by a planet changes at every point on the orbit B The time taken to complete an orbit is proportional to the $1.5th$ power of the radius of the orbit C Kepler’s 2nd law is a consequence of the conservation of angular momentum in central forces D The angular momentum of the planet changes at every point on the orbit E Kepler’s 2nd law is true only for circular orbits
Question 11 Explanation:
Option A $→$ False, because the area swept per unit time is a constant

Option B $→$ False, because the 2nd law of Kepler is based on rate of change of area

Option C $→$ True, because the 2nd law of Kepler can be derived using law of conservation of angular momentum

Option D $→$ False, because the angular momentum of the planet is a constant

Option E $→$ False, because it is true for circular and elliptical orbits
 Question 12

### What should be the perihelion speed of planet $A$?

 A $1.5v_A \ \cos⁡ θ$ B $0.5v_A \ \sin ⁡θ$ C $2v_A \ \cos ⁡θ$ D $2v_A \ \sin⁡ θ$ E $3.5v_A \ \tan⁡ θ$
Question 12 Explanation:
At point shown,

Angular momentum

$L=m_A v_A r \ \sin⁡ (π-θ)$

$=m_A v_A r \ \sin ⁡θ$

Perihelion speed is when the planet is closest to the Sun

At perihelion,

Angular momentum

$L=m_A v_A' \ (0.5r) \ \sin 90°$

$=0.5m_A v_A' r$

Thus, $v_A'=2v_A \ \sin⁡ θ$
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