2.6 Gravitational Force — Worked Solutions
Work through Topic 2.6
These are the full worked solutions to the lesson questions. Try the questions first, then check each step and the reason for it.
Question 1
At twice the distance from a planet center, what is the field strength?
- Half as large
- One quarter as large
- Twice as large
Step 1. For the same planet, write g_old = GM/r² and g_new = GM/(2r)².
Why this step? The central mass and G are unchanged; the distance is explicitly measured from the center, where the inverse-square model applies.
Step 2. Take the ratio g_new/g_old = r²/(4r²) = 1/4.
Why this step? Squaring the doubled distance produces a factor of four in the denominator, not a factor of two.
Step 3. Choose one quarter as large.
Why this step? The field is an inverse-square quantity; a half-size answer would instead describe an inverse-first-power dependence.
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Question 2
An elevator moves upward but slows down. Compared with mg, a passenger’s apparent weight is…
- Larger
- Smaller
- Always equal
Step 1. Upward motion that is slowing has downward acceleration, so a_y < 0 with upward positive.
Why this step? The acceleration sign follows the velocity change; slowing an upward velocity means subtracting from it.
Step 2. Use N − mg = ma_y, giving N = mg + ma_y < mg.
Why this step? A downward resultant requires weight to exceed the upward support force while the passenger remains supported.
Step 3. The apparent weight is smaller than mg.
Why this step? The scale measures N; upward velocity alone does not make the scale read larger because the force equation uses acceleration.
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Question 3
Why can an orbiting astronaut be apparently weightless?
- Gravity is absent in orbit
- Gravity is the only significant force, so no support normal force is needed
- The astronaut has no inertial mass
Step 1. An orbiting astronaut and spacecraft are both accelerating inward under gravity.
Why this step? Orbit is ongoing free fall, not a force-free straight-line motion; gravity supplies the changing velocity direction.
Step 2. No floor force is needed to hold the astronaut stationary relative to the freely falling craft, so the support normal force can be zero.
Why this step? Astronaut and craft can fall together rather than the floor opposing gravity as it does for a stationary person on Earth.
Step 3. Apparent weight can therefore vanish while gravitational force and inertial mass remain nonzero.
Why this step? A support-force reading defines apparent weight; interpreting its zero as absent gravity would remove the force needed for the orbit.
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Question 4
A 50 kg rider has upward acceleration 1.5 m/s². Find weight and scale reading using g = 10 m/s². Then find the scale reading for downward acceleration of the same magnitude.
Step 1. Compute gravitational weight: mg = (50 kg)(10 m/s²) = 500 N downward.
Why this step? The stated mass and local field determine weight independently of elevator acceleration.
Step 2. For upward acceleration, N − 500 = 50(+1.5), so N = 575 N.
Why this step? The support must exceed gravity by 75 N to create the specified upward resultant.
Step 3. For downward acceleration, N − 500 = 50(−1.5), so N = 425 N.
Why this step? Keep the same upward-positive axis; the negative acceleration now requires the support to be 75 N less than gravity.
Step 4. Weight stays 500 N; only the scale reading changes.
Why this step? Neither elevator case changes the mass or local gravitational field, but each demands a different contact force.
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Question 5
Derive the ratio of gravitational forces when both masses double and center separation triples. State why the same ratio does not apply if the separation is merely a surface altitude.
Step 1. Write F_old = Gm₁m₂/r² and F_new = G(2m₁)(2m₂)/(3r)².
Why this step? Universal gravitation multiplies the two masses and squares center separation, so each changed quantity must appear in its own factor.
Step 2. Divide: F_new/F_old = 4/9.
Why this step? Both mass doublings supply a factor of four; tripling separation supplies a factor of nine that partly offsets it.
Step 3. If h is surface altitude instead, center separation is R + h, not h.
Why this step? A planet’s gravitational model is centered at its mass center; tripling h does not generally triple R + h.
Step 4. For fixed radius R, changing h to 3h would instead give the distance factor (R + h)²/(R + 3h)².
Why this step? Keeping the planet radius in the ratio prevents misapplying the 1/9 distance factor to an altitude change.
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Question 6
Sketch gravitational field strength versus center distance outside a spherical planet. Mark R and 2R and their field values. Explain how a short near-surface interval can look approximately constant.
Step 1. For r ≥ R, sketch the magnitude g = GM/r² as a positive decreasing curve.
Why this step? Outside a spherical source, attraction gives an inward vector field whose magnitude follows this center-distance dependence.
Step 2. Mark g(R) = g₀ and g(2R) = g₀/4.
Why this step? The two labeled distances provide a quantitative scale check on the inverse-square graph.
Step 3. For a small height h, g(R + h)/g₀ = 1/(1 + h/R)² is close to one when h/R is tiny.
Why this step? Near-surface classroom changes are small compared with the whole planet radius, so the relative denominator change is small.
Step 4. Treat g as constant only over that short interval, not over the full R-to-2R range.
Why this step? The quarter-strength point shows that extending the approximation to planet-scale distances would hide a substantial change.