Astronauts on the International Space Station appear weightless because

Answer: They are in free fall (continuously falling around Earth). They orbit because gravity pulls them — they're in continuous free fall. Apparent weightlessness arises from the lack of normal force on them, not from absence of gravity.

The astronaut's <strongweight</strong on the Moon is closest to

Answer: 112\text{N} W = m gM = (70)(1.6) = 112\text{N}.

The acceleration due to Earth's gravity at altitude h above the surface (Earth radius RE) is

Answer: gE\,(RE/(RE + h))^2 From g = G M/r^2 with r = RE + h: g(h) = gE\,(RE/(RE + h))^2.

AP Physics 1 · Topic 2.6

Gravitational and Electric Forces Practice

Q: The astronaut's <strongweight</strong on the Moon is closest to

Answer: 112\text{N} W = m gM = (70)(1.6) = 112\text{N}.

Q: The acceleration due to Earth's gravity at altitude h above the surface (Earth radius RE) is

Answer: gE\,(RE/(RE + h))^2 From g = G M/r^2 with r = RE + h: g(h) = gE\,(RE/(RE + h))^2.

Part of Force and Translational Dynamics.(TOP-2.F)

Practice questions

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Sample questions

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Sample 1difficulty 1/5
$Astronauts on the International Space Station appear weightless because$
- A
  $They are in free fall (continuously falling around Earth).$
  check_circle
- B
  $There is no gravity at that altitude.$
- C
  $Their masses become zero in orbit.$
- D
  $They are above the atmosphere.$
Why
$They orbit because gravity pulls them — they're in continuous free fall. Apparent weightlessness arises from the lack of normal force on them, not from absence of gravity.$
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Sample 2difficulty 1/5
$The astronaut's <strong> weight </strong> on the Moon is closest to$
- A
  $11~\text{N}$
- B
  $700~\text{N}$
- C
  $112~\text{N}$
  check_circle
- D
  $1120~\text{N}$
Why
$W = m g_M = (70)(1.6) = 112~\text{N}$
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Sample 3difficulty 2/5
$h$
- A
  $g_E\,(R_E/(R_E + h))^2$
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- B
  $g_E$
- C
  $g_E \, R_E/(R_E + h)$
- D
  $g_E\,(h/R_E)^2$
Why
$g = G M/r^2$
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Sample 4difficulty 3/5
$A satellite in circular orbit is held by gravity, which provides the centripetal force. Doubling the orbital radius (same Earth) changes the satellite's orbital speed by a factor of$
- A
  $1/2$
- B
  $1/\sqrt{2}$
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- C
  $2$
- D
  $\sqrt{2}$
Why
$G M m/r^2 = m v^2/r \Rightarrow v = \sqrt{G M/r}$
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Sample 5difficulty 4/5
$g$
- A
  $g_X = (3/2) g$
- B
  $g_X = (4/3) g$
- C
  $g_X = (3/4) g$
  check_circle
- D
  $g_X = (1/4) g$
Why
$g \propto M/R^2$
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