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

AP Physics C: Mechanics practice test 7.

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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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