Fields and gravity

A satellite moves in a circular orbit around Earth at constant speed. Which statement best describes the force on it?

1. A

   No force acts because its speed is constant

2. B

   An outward force balances gravity

3. C

   Gravity provides a resultant force towards Earth's centre

4. D

   The satellite's engine continuously pushes it forwards

1. a

   Select and explain Select the correct statement and explain why constant speed does not mean zero acceleration.

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   Correct choiceC

   1. 1

      Selects option C: gravity provides the resultant force directed towards Earth's centre.

   2. 2

      States that velocity changes because its direction changes even though speed is constant.

   3. 3

      Links the inward acceleration to an inward resultant gravitational force.

At Earth's surface the gravitational field strength is 9.8 N kg⁻¹. A 500 kg satellite is at a distance of two Earth radii from Earth's centre. Treat Earth as a sphere and use the inverse-square relationship.

1. a

   Calculate Calculate the gravitational field strength at the satellite.

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   1. 1

      Recognises that doubling distance from the centre reduces field strength by 2² = 4.

   2. 2

      Calculates 9.8 ÷ 4.

   3. 3

      Obtains 2.45 N kg⁻¹.

2. b

   Calculate Calculate the gravitational force on the satellite.

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   1. 1

      Uses F = mg or F = mass × field strength.

   2. 2

      Obtains F = 500 × 2.45 = 1225 N, towards Earth's centre.

3. c

   Explain Explain why the satellite is not weightless at this position.

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   1. 1

      Gravity is still acting; apparent weightlessness in orbit is due to continuous free fall, not zero gravitational field.

A simulation reports the electric field magnitude at different distances from one isolated positive point charge.

1. a

   Show Use two data pairs to show that the field follows an inverse-square pattern.

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   1. 1

      Selects two suitable distances, such as 0.10 m and 0.20 m.

   2. 2

      Shows that distance doubles while field falls from 360 to 90 N C⁻¹, a factor of four.

   3. 3

      Connects a factor-four decrease to 1 ÷ 2², supporting E ∝ 1/r².

2. b

   Predict Predict the field magnitude at 0.50 m.

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   1. 1

      Uses the constant Er² = 360 × 0.10² = 3.6 N m² C⁻¹.

   2. 2

      Calculates E = 3.6 ÷ 0.50² = 14.4 N C⁻¹.

3. c

   Evaluate State two reasons why a classroom field-line diagram is not a literal picture of this field.

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   1. 1

      States that the lines are a chosen representation and there are not physical threads in space.

   2. 2

      States another limitation, such as line number or spacing being chosen, a flat drawing representing a three-dimensional field, or arrows showing test-charge direction rather than motion.