Calculate Gravitational Force Mathematically - Grade 11 Physics

1. Planet P has radius $R$ and mass $M$. Planet Q has radius $2R$ and the same average density as planet P. An object experiences a gravitational force of $90\,\mathrm{N}$ at the surface of planet P. What force would the same object experience at an altitude of $R$ above the surface of planet Q?
   Answer: $80\,\mathrm{N}$, accounting for both the larger mass and the center distance Whenever you see a gravitational force problem involving two planets with the same density but different sizes, you need to track two things simultaneously: how the mass changes with radius, and how the distance to the center changes. Start with the relationship between density, mass, and radius. Since density $\rho = \frac{M}{\frac{4}{3}\pi R^3}$ is the same for both planets, and planet Q has radius $2R$, its mass scales as $R^3$: $M_Q = 8M$. Now apply Newton's law of gravitation, $F = \frac{GMm}{d^2}$, at altitude $R$ above planet Q's surface. The distance from Q's center is $2R + R = 3R$. So the force on planet Q becomes: $F_Q = \frac{G(8M)m}{(3R)^2} = \frac{8GMm}{9R^2}$ On planet P's surface: $F_P = \frac{GMm}{R^2} = 90\,\mathrm{N}$. Therefore: $F_Q = \frac{8}{9} \times 90\,\mathrm{N} = 80\,\mathrm{N}$ This confirms B is correct. A ignores the altitude entirely, computing gravity at Q's surface (which gives $2 \times 90 = 180\,\mathrm{N}$ since surface gravity scales as $M/R^2 \propto R$) — a classic "forgot the altitude" error. C uses the eightfold mass increase but forgets to divide by the squared distance, producing an inflated answer. D applies an inverse-square factor of 9 (as if the distance were $3R$ from planet P's center) but uses planet P's original mass, ignoring Q's larger mass entirely. Your strategy: always write out $F = \frac{GMm}{d^2}$ explicitly, substituting the center-to-object distance — not just the radius or altitude alone.
2. A $200\,\mathrm{kg}$ spacecraft is attracted by two small moons. One moon, due east of the spacecraft, has mass $6.00\times10^{20}\,\mathrm{kg}$ and is $2.00\times10^5\,\mathrm{m}$ away. The other, due north, has mass $3.00\times10^{20}\,\mathrm{kg}$ and is $1.00\times10^5\,\mathrm{m}$ away. What is the approximate net gravitational force on the spacecraft?
   Answer: $447\,\mathrm{N}$ directed $63.4^\circ$ north of east, from vector addition When a problem involves forces pulling in perpendicular directions, you must treat them as vectors — calculate each force separately, then combine them using vector addition (the Pythagorean theorem), never by simply adding or subtracting magnitudes. Start with Newton's Law of Gravitation: $F = \frac{Gm_1m_2}{r^2}$, where $G = 6.67\times10^{-11}\,\mathrm{N\cdot m^2/kg^2}$. Force from the eastern moon (F₁, pointing east): $F_1 = \frac{(6.67\times10^{-11})(200)(6.00\times10^{20})}{(2.00\times10^5)^2} = \frac{8.004\times10^{12}}{4.00\times10^{10}} \approx 200\,\mathrm{N}$ Force from the northern moon (F₂, pointing north): $F_2 = \frac{(6.67\times10^{-11})(200)(3.00\times10^{20})}{(1.00\times10^5)^2} = \frac{4.002\times10^{12}}{1.00\times10^{10}} \approx 400\,\mathrm{N}$ Net force magnitude: $F_{net} = \sqrt{200^2 + 400^2} = \sqrt{200000} \approx 447\,\mathrm{N}$ Direction: $\theta = \tan^{-1}\!\left(\frac{F_2}{F_1}\right) = \tan^{-1}(2) \approx 63.4^\circ$ north of east. This confirms C. Choice A incorrectly adds the magnitudes directly ($200 + 400 = 600\,\mathrm{N}$), ignoring that perpendicular vectors require the Pythagorean theorem. Choice B gets the correct magnitude but reverses which component is east vs. north, swapping the angle to $26.6°$. Choice D subtracts the forces, which has no physical basis here. Study tip: Whenever forces act at right angles, draw a quick sketch, label each component, and always use $\sqrt{F_x^2 + F_y^2}$ for the magnitude — never add or subtract them directly.
3. A planet has twice Earth's mass and twice its radius. The surface gravitational force, compared with Earth's, is
   Answer: half as large Surface gravity depends on mass divided by the square of the radius. Doubling the mass doubles the force, but doubling the radius quadruples the denominator, so the net force is 2 / 4 = 1/2 of Earth's. The tempting wrong answer is thinking only the mass matters, which would make it twice as large.
4. An astronaut weighs $720\,\mathrm{N}$ at Earth's surface. Earth is modeled as a sphere of radius $R$. What is the astronaut's gravitational weight at an altitude of $R/2$ above the surface?
   Answer: $320\,\mathrm{N}$, because the center-to-center distance is $1.5R$ Whenever you see a question about gravity at different altitudes, the key concept is Newton's Law of Universal Gravitation: gravitational force falls off as the square of the distance between centers — not the distance from the surface. At Earth's surface, the center-to-center distance is simply $R$. When the astronaut rises to an altitude of $R/2$ above the surface, the new center-to-center distance becomes $R + R/2 = 1.5R$. The ratio of the new distance to the original is $1.5$, so the force scales by $\frac{1}{(1.5)^2} = \frac{1}{2.25}$. Multiplying: $720 \times \frac{1}{2.25} = 320\,\mathrm{N}$. That confirms C is correct. Choice A claims the astronaut is "twice as far from Earth," which misidentifies the new distance. The center-to-center distance goes from $R$ to $1.5R$, not $2R$ — and even if it were $2R$, the force would be $\frac{1}{4}$ of 720 N = 180 N only at that doubling. A is both using the wrong distance and forgetting the square. Choice B says force decreases linearly with distance — this is flat-out wrong. Gravity follows an inverse-square law, not an inverse law. Linear thinking leads to completely incorrect results in orbital and altitude problems. Choice D gets the inverse-square idea right but misapplies it, inflating rather than reducing the force — suggesting a calculation error in the distance ratio. Study tip: Always measure center-to-center distance (not surface-to-surface), then square the ratio. Writing $F \propto \frac{1}{d^2}$ before plugging in numbers will keep you from falling into the linear-distance trap.
5. Two spherical asteroids have masses of $2.00\times10^4\,\mathrm{kg}$ and $5.00\times10^4\,\mathrm{kg}$ and radii of $2.00\,\mathrm{m}$ and $3.00\,\mathrm{m}$, respectively. Their surfaces are separated by $4.00\,\mathrm{m}$. Using $G=6.67\times10^{-11}\,\mathrm{N\,m^2/kg^2}$, what gravitational force do they exert on each other?
   Answer: $8.23\times10^{-4}\,\mathrm{N}$, using the full center-to-center separation When applying Newton's Law of Universal Gravitation, the separation distance r must always be measured center-to-center, not surface-to-surface. This is because the law treats spherical masses as point masses located at their geometric centers. Here, the center-to-center distance is the sum of both radii plus the surface gap: $r = 2.00\,\mathrm{m} + 3.00\,\mathrm{m} + 4.00\,\mathrm{m} = 9.00\,\mathrm{m}$ Plugging into $F = \frac{Gm_1m_2}{r^2}$: $F = \frac{(6.67\times10^{-11})(2.00\times10^4)(5.00\times10^4)}{(9.00)^2} = \frac{6.67\times10^{-2}}{81} \approx 8.23\times10^{-4}\,\mathrm{N}$ This confirms D is correct. Each wrong answer reflects a specific mistake in computing r. Choice A uses only the 4.00 m surface gap as the separation, ignoring both radii entirely — this dramatically underestimates r and inflates the force. Choice B uses only the sum of the two radii (5.00 m), forgetting to include the 4.00 m gap between the surfaces — another incomplete accounting of the total distance. Choice C includes one radius and the gap (say, 3.00 + 4.00 = 7.00 m) but omits the other radius — a partial but still incorrect center-to-center distance. A reliable strategy: whenever a problem gives you surface separation and two radii, always reconstruct the full center-to-center distance as $r = r_1 + d_{\text{gap}} + r_2$ before substituting into the gravitational force formula. Skipping any piece of this sum is the most common trap on gravitation problems.
6. A 2.0 kg mass and a 0.50 kg mass are 50 cm apart. Find the gravitational force.
   Answer: $2.7\times10^{-10}$ N Convert 50 cm to 0.50 m and square it: $(0.50)^2$ = 0.25. Multiply G = 6.67e-11 by 2.0 kg and 0.50 kg, then divide by 0.25, giving 2.7e-10 N. The tempting wrong answer comes from using 0.50 without squaring, which gives 1.3e-10 N, but the distance must be squared in Newton's law.
7. At Earth's surface an object weighs 900 N. At a height 2R above the surface, its weight is? (R = Earth's radius)
   Answer: 100 N Gravity follows an inverse-square law with distance from Earth's center. At the surface, distance is R; at height 2R above the surface, distance is R + 2R = 3R. So weight becomes 900/9 = 100 N. The tempting mistake is to treat 2R as the new distance and divide 900 by 4, giving 225 N, but you must include the Earth's radius as part of the distance.
8. A $500\,\mathrm{kg}$ satellite travels in a nearly circular orbit with orbital radius $7.00\times10^6\,\mathrm{m}$ and period $6.00\times10^3\,\mathrm{s}$. Assuming gravity provides the entire centripetal force, what is the gravitational force on the satellite?
   Answer: $3.84\times10^3\,\mathrm{N}$, by applying the centripetal-force model with orbital speed When a satellite orbits in a circle, gravity is the centripetal force — so you can set $F_g = \frac{mv^2}{r}$. The cleanest approach uses orbital speed: since the satellite travels a circumference $2\pi r$ in one period $T$, its speed is $v = \frac{2\pi r}{T}$. Substituting gives the standard formula: $F_g = \frac{4\pi^2 m r}{T^2}$ Plugging in: $m = 500\,\text{kg}$, $r = 7.00\times10^6\,\text{m}$, $T = 6.00\times10^3\,\text{s}$: $F_g = \frac{4\pi^2 (500)(7.00\times10^6)}{(6.00\times10^3)^2} \approx 3.84\times10^3\,\text{N}$ This confirms C is correct. Choice A arrives at $97.2\,\text{N}$ by writing $F = mr/T^2$ — it forgets the factor $4\pi^2$, which comes from squaring the $2\pi$ in the circumference. Dropping it massively underestimates the force. Choice B gets $1.92\times10^3\,\text{N}$ by mistaking the orbital radius for the diameter, effectively halving $r$ before computing. The radius is already the distance from center to satellite — no further halving needed. Choice D doubles the correct answer to $7.68\times10^3\,\text{N}$ by using $8\pi^2$ instead of $4\pi^2$. This is a pure arithmetic error with no physical justification. Study tip: Memorize $v = \frac{2\pi r}{T}$ and $F_c = \frac{mv^2}{r}$ as your two building blocks for circular orbit problems. Combining them always yields $\frac{4\pi^2 mr}{T^2}$ — if your answer doesn't have that $4\pi^2$, something went wrong.
9. A $1.00\times10^3\,\mathrm{kg}$ probe lies on the line between asteroid A and asteroid B. Asteroid A has mass $8.00\times10^{22}\,\mathrm{kg}$ and is $2.00\times10^7\,\mathrm{m}$ from the probe. Asteroid B has mass $2.00\times10^{22}\,\mathrm{kg}$ and is $1.00\times10^7\,\mathrm{m}$ from the probe. What is the net gravitational force on the probe?
   Answer: $0\,\mathrm{N}$, because the two gravitational forces have equal magnitudes and opposite directions When a probe sits between two massive bodies, you need to compare the gravitational force each body exerts and determine whether they reinforce or oppose each other. Use Newton's law of gravitation: $F = \frac{Gm_1m_2}{r^2}$, where $G = 6.67\times10^{-11}\,\mathrm{N\cdot m^2/kg^2}$. Force from asteroid A (mass $8.00\times10^{22}\,\mathrm{kg}$, distance $2.00\times10^7\,\mathrm{m}$): $F_A = \frac{(6.67\times10^{-11})(8.00\times10^{22})(1.00\times10^3)}{(2.00\times10^7)^2} = \frac{5.336\times10^{16}}{4.00\times10^{14}} \approx 133\,\mathrm{N}$ Wait — let me redo carefully: $F_A = \frac{(6.67\times10^{-11})(8.00\times10^{22})(10^3)}{4\times10^{14}} = \frac{5.336\times10^{15}}{4\times10^{14}} \approx 13.3\,\mathrm{N toward A}$ Force from asteroid B (mass $2.00\times10^{22}\,\mathrm{kg}$, distance $1.00\times10^7\,\mathrm{m}$): $F_B = \frac{(6.67\times10^{-11})(2.00\times10^{22})(10^3)}{(10^7)^2} = \frac{1.334\times10^{15}}{10^{14}} \approx 13.3\,\mathrm{N toward B}$ Both forces are equal in magnitude (≈13.3 N) but point in opposite directions, so they cancel — net force is 0 N, confirming answer A. Choice B is tempting because asteroid A is more massive, but greater mass doesn't automatically win — distance matters too, and here the closer distance of B exactly compensates. Choice C incorrectly adds the magnitudes as if both forces point the same way. Choice D correctly identifies that closeness matters but draws the wrong conclusion. Study tip: Always calculate both forces before assuming which "wins." On these problems, mass and distance effects can perfectly cancel — never assume the heavier or closer body automatically dominates.
10. Two objects initially exert a gravitational force of magnitude $F$ on each other. The first object's mass is tripled, the second object's mass is halved, and their center-to-center separation is decreased by $25\%$. What is the new gravitational force?
    Answer: $\dfrac{8}{3}F$, using the changed mass product and inverse-square distance factor Whenever you see a gravitational force problem where multiple quantities change simultaneously, your strategy is to track each change as a multiplicative factor applied to Newton's Law of Gravitation: $F = G\dfrac{m_1 m_2}{r^2}$. Start with the original force $F = G\dfrac{m_1 m_2}{r^2}$. Now apply each change: - First mass tripled: $m_1 \to 3m_1$ - Second mass halved: $m_2 \to \frac{1}{2}m_2$ - Separation decreased by 25%: $r \to 0.75r = \frac{3}{4}r$ The new force becomes: $F' = G\dfrac{(3m_1)\!\left(\frac{1}{2}m_2\right)}{\left(\frac{3}{4}r\right)^2} = G\dfrac{\frac{3}{2}m_1 m_2}{\frac{9}{16}r^2} = \frac{3}{2} \cdot \frac{16}{9} \cdot G\dfrac{m_1 m_2}{r^2} = \frac{8}{3}F$ Answer B is correct for exactly this reason — it properly applies the inverse-square to the new separation. Answer A is wrong because it treats distance with an inverse-*linear* factor $\left(\frac{1}{r}\right)$ rather than inverse-square $\left(\frac{1}{r^2}\right)$, which contradicts Newton's law entirely. Answer C is wrong because it ignores the distance change completely, only multiplying the mass factor $\frac{3}{2}$ without accounting for the decreased separation — a common oversight. Answer D is wrong because it multiplies by $\left(\frac{3}{4}\right)^2$ instead of dividing by it, getting the direction of the distance effect backwards. Study tip: Always write out the full ratio $\dfrac{F'}{F}$ and substitute each changed quantity explicitly — this prevents you from accidentally dropping a factor or misapplying the inverse-square relationship.
11. A $250\,\mathrm{kg}$ spacecraft experiences a gravitational force of $2.45\times10^3\,\mathrm{N}$ at Earth's surface. At what altitude will its gravitational force be $612.5\,\mathrm{N}$? Use an Earth radius of $6.40\times10^6\,\mathrm{m}$.
    Answer: $6.40\times10^6\,\mathrm{m}$ above the surface, where the center distance is $2R$ Whenever you see a question about gravity changing with distance, your first instinct should be Newton's inverse-square law: gravitational force drops off as $\frac{1}{r^2}$, where $r$ is measured from Earth's center, not the surface. Start by finding the force ratio. The force drops from $2.45\times10^3\,\mathrm{N}$ to $612.5\,\mathrm{N}$, which is exactly one-quarter: $\frac{612.5}{2450} = \frac{1}{4}$. Since $F \propto \frac{1}{r^2}$, if the force becomes $\frac{1}{4}$ as large, the distance must become $\sqrt{4} = 2$ times as large. So the new center distance is $2R = 2\times(6.40\times10^6) = 1.28\times10^7\,\mathrm{m}$. The altitude (above the surface) is $2R - R = R = 6.40\times10^6\,\mathrm{m}$. That confirms A. Choice B makes the critical error of reporting the center distance ($1.28\times10^7\,\mathrm{m}$) as the altitude, forgetting to subtract Earth's radius. Choice C applies an inverse-linear relationship ($F \propto \frac{1}{r}$) instead of the correct inverse-square law, producing an answer that's triple the correct altitude. Choice D incorrectly reasons that halving the radius gives one-fourth the force — this confuses linear and squared scaling, and also moves the spacecraft closer to Earth, which would increase force, not decrease it. A reliable strategy: always convert the force ratio into a distance ratio using $\frac{F_1}{F_2} = \frac{r_2^2}{r_1^2}$, solve for the new center distance, then subtract $R$ to get altitude.
12. A spherical planet has radius $4.00\times10^6\,\mathrm{m}$, and the gravitational field strength at its surface is $12.0\,\mathrm{N/kg}$. What gravitational force acts on a $50.0\,\mathrm{kg}$ probe at an altitude of $2.00\times10^6\,\mathrm{m}$ above the surface?
    Answer: $267\,\mathrm{N}$, using the total distance from the planet's center Gravity questions involving altitude test your understanding of the inverse-square law: gravitational field strength decreases with the square of the distance from the planet's center, not from the surface. The critical setup step is always measuring distance from the center, not the surface. Here, the planet's radius is $R = 4.00\times10^6\,\mathrm{m}$ and the probe sits $2.00\times10^6\,\mathrm{m}$ above the surface, so its total distance from the center is $r = 4.00\times10^6 + 2.00\times10^6 = 6.00\times10^6\,\mathrm{m}$. The field scales as $g' = g_s\left(\frac{R}{r}\right)^2 = 12.0\times\left(\frac{4.00}{6.00}\right)^2 = 12.0\times\frac{4}{9} \approx 5.33\,\mathrm{N/kg}$. The gravitational force on the probe is then $F = mg' = 50.0\times5.33 \approx 267\,\mathrm{N}$, confirming B. Choice A is wrong because gravity follows an inverse-*square* law, not an inverse-linear one — halving the ratio of distances would give a different (incorrect) result. Choice C misidentifies the probe's distance from the center as $2R$ (i.e., $8.00\times10^6\,\mathrm{m}$), forgetting to add the planet's own radius to the altitude. Choice D inverts the distance ratio incorrectly, using $(r/R)^2$ instead of $(R/r)^2$, which amplifies gravity instead of reducing it. Your go-to strategy: always convert altitude to center-to-center distance first, then apply $g' = g_s(R/r)^2$. Writing that ratio down before plugging in numbers will save you from the most common traps here.
13. Two spheres have gravitational force $4.0\times10^{-11}$ N at 2.0 m. What is the force at 0.50 m?
    Answer: $6.4\times10^{-10}$ N Gravitational force follows an inverse-square law, so when distance shrinks from 2.0 m to 0.50 m, it shrinks by a factor of 4. That means the force increases by 4 squared, or 16 times: 4.0e-11 N times 16 = 6.4e-10 N. A tempting wrong answer is 1.6e-10 N from multiplying by 4 only, but you must square the distance ratio.
14. A laboratory model treats two small objects as point masses. A $150\,\mathrm{kg}$ object and an unknown object are $5.00\,\mathrm{m}$ apart and attract each other with a force of $2.00\times10^{-7}\,\mathrm{N}$. Using $G=6.67\times10^{-11}\,\mathrm{N\,m^2/kg^2}$, what is the unknown mass?
    Answer: $5.00\times10^2\,\mathrm{kg}$, after applying the inverse-square force model Whenever you see a gravitational force problem, your go-to tool is Newton's Law of Universal Gravitation: $F = \frac{G m_1 m_2}{r^2}$ The key feature here is the inverse-square relationship — the separation distance must be squared. To find the unknown mass, rearrange for $m_2$: $m_2 = \frac{F r^2}{G m_1}$ Plugging in the values: $m_2 = \frac{(2.00\times10^{-7})(5.00)^2}{(6.67\times10^{-11})(150)}$ $m_2 = \frac{(2.00\times10^{-7})(25.0)}{1.00\times10^{-8}} = \frac{5.00\times10^{-6}}{1.00\times10^{-8}} = 5.00\times10^{2}\,\mathrm{kg}$ This confirms D is correct — it properly applies the inverse-square model. Choice A ($50.0\,\mathrm{kg}$) results from a powers-of-ten arithmetic error during simplification, likely mishandling the exponents when dividing. Choice B ($100\,\mathrm{kg}$) comes from using $r = 5.00$ instead of $r^2 = 25.0$ — a very common mistake where students forget to square the separation before substituting. Choice C ($2.00\times10^3\,\mathrm{kg}$) likely arises from misinterpreting "separation" as a radius in some circular-motion context and applying an incorrect formula altogether. The single most important habit for these problems: **always write $r^2$ explicitly before substituting numbers.** If you write the formula out fully before plugging anything in, you dramatically reduce the chance of forgetting to square the distance — which is the trap this question is specifically designed to test.
15. Two spheres attract with force F. If both masses double and distance triples, the new force is
    Answer: $\frac{4}{9}F$ Doubling both masses multiplies the gravitational force by 4 because each mass doubles. Tripling the distance divides the force by 3 squared = 9, so the new force is 4/9 F. The tempting error is dividing by only 3 for the distance, giving 4/3 F, but the inverse-square law means the distance factor must be squared.

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