Physics Flashcards: Calculate Gravitational Force Mathematically

Study Calculate Gravitational Force Mathematically in Physics with focused flashcards that help you recognize the idea, recall the key rule, and apply it in practice-style prompts.

Physics

Calculate Gravitational Force Mathematically

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QUESTION
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What happens to FF if the distance rr between two masses is doubled?

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ANSWER

F becomes 14 as largeF\text{ becomes }\frac{1}{4}\text{ as large}. Force varies as 1r2\frac{1}{r^2}, so (2r)2=4(2r)^2=4 in denominator.

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Flashcard 1: What happens to FF if the distance rr between two masses is doubled?

Answer: F becomes 14 as largeF\text{ becomes }\frac{1}{4}\text{ as large}. Force varies as 1r2\frac{1}{r^2}, so (2r)2=4(2r)^2=4 in denominator.

Flashcard 2: Find m2m_2 in terms of FF using F=Gm1m2r2F=G\frac{m_1m_2}{r^2}.

Answer: m2=Fr2Gm1m_2=\frac{Fr^2}{Gm_1}. Isolate m2m_2 by multiplying both sides by r2/(Gm1)r^2/(Gm_1).

Flashcard 3: Calculate weight WW for m=2 kgm=2\ \text{kg} in a field g=10 m/s2g=10\ \text{m/s}^2.

Answer: 20 N20\ \text{N}. Direct multiplication: W=2×10=20W=2\times10=20.

Flashcard 4: State the relationship between gravitational force and field strength for a test mass mm.

Answer: F=mgF=mg. Force equals mass times gravitational field strength.

Flashcard 5: What happens to FF if the distance rr between two masses is halved?

Answer: F becomes 4 times as largeF\text{ becomes }4\text{ times as large}. Force varies as 1r2\frac{1}{r^2}, so (r2)2=14(\frac{r}{2})^2=\frac{1}{4} in denominator.

Flashcard 6: What happens to FF if the separation distance doubles from rr to 2r2r?

Answer: FF4F\to\frac{F}{4}. Force varies as 1r2\frac{1}{r^2}, so doubling rr gives 14\frac{1}{4} the force.

Flashcard 7: Identify the SI unit of gravitational force FF.

Answer: N\text{N}. Force is measured in newtons in the SI system.

Flashcard 8: State the formula for gravitational potential energy of two masses separated by distance rr.

Answer: U=Gm1m2rU=-G\frac{m_1m_2}{r}. Negative sign indicates attractive force; zero at infinite separation.

Flashcard 9: What happens to FF if the separation distance is halved from rr to r2\frac{r}{2}?

Answer: F4FF\to^4F. Halving distance means r2r24r^2\to\frac{r^2}{4}, so force quadruples.

Flashcard 10: What value of the gravitational constant should you use in SI calculations?

Answer: G=6.67×1011 Nm2kg2G=6.67\times10^{-11}\ \text{N}\cdot\text{m}^2\cdot\text{kg}^{-2}. Standard value used in SI unit calculations.

Flashcard 11: State the formula for the magnitude of gravitational force between two point masses.

Answer: F=Gm1m2r2F=G\frac{m_1m_2}{r^2}. Newton's law of universal gravitation relates force to masses and inverse square of distance.

Flashcard 12: Identify the meaning of rr in F=Gm1m2r2F=G\frac{m_1m_2}{r^2}.

Answer: r=center-to-center distance between the massesr=\text{center-to-center distance between the masses}. Distance measured from center of mass to center of mass.

Flashcard 13: Identify the SI unit of mass mm used in Newton's law of gravitation.

Answer: kg\text{kg}. Mass is measured in kilograms in SI units.

Flashcard 14: Identify the variable in F=Gm1m2r2F=G\frac{m_1m_2}{r^2} that represents center-to-center separation.

Answer: rr. Distance between centers of the two masses.

Flashcard 15: What is the ratio F2F1\frac{F_2}{F_1} if both masses double and distance stays the same?

Answer: 44. Doubling both masses gives 2×2=42\times^2=4 times the force.

Flashcard 16: What is the SI unit of the gravitational constant GG?

Answer: Nm2kg2\text{N}\cdot\text{m}^2\cdot\text{kg}^{-2}. Derived from F=maF=ma and dimensional analysis of the formula.

Flashcard 17: Calculate FF for m1=2 kgm_1=2\ \text{kg}, m2=3 kgm_2=3\ \text{kg}, r=1 mr=1\ \text{m}.

Answer: F4.00×1010 NF\approx^4.00\times10^{-10}\ \text{N}. Calculate: F=6.67×1011×2×3/12=4.00×1010F=6.67×10^{-11}×2×3/1^2=4.00×10^{-10} N.

Flashcard 18: What happens to FF if one mass doubles from m1m_1 to 2m12m_1 (all else constant)?

Answer: F2FF\to^2F. Force is directly proportional to each mass.

Flashcard 19: Find rr in terms of FF using F=Gm1m2r2F=G\frac{m_1m_2}{r^2}.

Answer: r=Gm1m2Fr=\sqrt{G\frac{m_1m_2}{F}}. Solve for rr by taking square root of rearranged equation.

Flashcard 20: Calculate FF for m1=1 kgm_1=1\ \text{kg}, m2=1 kgm_2=1\ \text{kg}, r=2 mr=2\ \text{m}.

Answer: F1.67×1011 NF\approx^1.67\times10^{-11}\ \text{N}. Calculate: F=6.67×1011×1×1/22=1.67×1011F=6.67×10^{-11}×1×1/2^2=1.67×10^{-11} N.

Flashcard 21: State the formula for gravitational field strength gg at distance rr from mass MM.

Answer: g=GMr2g=G\frac{M}{r^2}. Field strength is force per unit mass at that location.

Flashcard 22: Identify the SI unit of distance rr used in F=Gm1m2r2F=G\frac{m_1m_2}{r^2}.

Answer: m\text{m}. Distance is measured in meters in SI units.

Flashcard 23: Calculate the ratio F2F1\frac{F_2}{F_1} if rr changes from r1r_1 to r2r_2 with masses constant.

Answer: F2F1=(r1r2)2\frac{F_2}{F_1}=\left(\frac{r_1}{r_2}\right)^2. Ratio of forces equals inverse ratio of distances squared.

Flashcard 24: What is the direction of the gravitational force on each mass in a two-body system?

Answer: Along the line joining the centers, toward the other mass\text{Along the line joining the centers, toward the other mass}. Gravity is always attractive between masses.

Flashcard 25: What is the ratio F2F1\frac{F_2}{F_1} if distance changes from rr to 3r3r (masses unchanged)?

Answer: 19\frac{1}{9}. Since F1r2F\propto\frac{1}{r^2}, tripling rr gives 19\frac{1}{9} the force.

Flashcard 26: What is the value of the universal gravitational constant GG in SI units?

Answer: G=6.67×1011 Nm2/kg2G=6.67\times10^{-11}\ \text{N}\cdot\text{m}^2/\text{kg}^2. Fundamental constant determined experimentally by Cavendish.

Flashcard 27: What happens to FF if one mass (for example m1m_1) is doubled while rr is constant?

Answer: F doublesF\text{ doubles}. Force is directly proportional to each mass.

Flashcard 28: Calculate gg at r=2Rr=2R from a planet: if g(R)=g0g(R)=g_0, what is g(2R)g(2R)?

Answer: g(2R)=g04g(2R)=\frac{g_0}{4}. Field follows inverse square law: g1/r2g∝1/r^2, so doubling rr gives g/4g/4.

Flashcard 29: Calculate FF on a 3 kg3\ \text{kg} mass where the gravitational field strength is g=10 ms2g=10\ \text{m}\cdot\text{s}^{-2}.

Answer: F=30 NF=30\ \text{N}. F=mg=3×10=30F=mg=3\times10=30 N using weight formula.

Flashcard 30: Which option gives the correct proportionality for Newtonian gravity with distance: FrF\propto r, F1rF\propto \frac{1}{r}, or F1r2F\propto \frac{1}{r^2}?

Answer: F1r2F\propto \frac{1}{r^2}. Inverse square law: force decreases with square of distance.

Flashcard 31: Identify the sign of gravitational potential energy UU for two isolated masses at finite rr.

Answer: U<0U<0. Attractive force means bound system has negative energy.

Flashcard 32: State the formula for gravitational field strength due to a mass MM at distance rr.

Answer: g=GMr2g=G\frac{M}{r^2}. Field strength is force per unit mass at distance rr.

Flashcard 33: State the relationship between weight WW and gravitational field strength gg for mass mm.

Answer: W=mgW=mg. Weight equals mass times gravitational field strength.