Gravity revision answers

 

1.       (a)     arrows (at least one) indicating direction is towards the planet.

B1

          All lines looking as though they would meet at the centre judged by eye

At least 4 drawn and care taken
Some of the lines must be outside the planet.

B1

  (b)   (i)      (mg = GMm/r2 and hence) M = gr2/G

Equation needs to be rearranged as shown for C1 mark

C1 M1

         correct substitution M = 24.9 × (7.14 × 107)2/6.67 × 10–11
                                 = 1.9 × 1027 Kg (i.e about 2 × 1027)

A1

  (ii)   correct substitution into V = (4/3)πr3 = (4/3)π(7.14 × 107)3
{= 1.52 × 1024 m3}

If m= 2 × 1027 kg is used

C1

         density = mass/volume = 1.9 × 1027/1.52 × 1024 = 1250 kg m–3

d = 1312 scores 2 marks

A1

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  2.     (a)     forces FS and FG acting inwards, force FE acting outwards - all through
centre of proton;
3 forces 2/2, 2 forces 1/2, marked and labelled (2)                                                2

  (b)   FE = FS + FG;
accept FE + FS + FG = 0 allow ecf from (a) (1)                                                     1

  (c)    (i)      FE = Q2 / (4π ε0 r2) (1)
= (1.6 × 10–19)2 / [4π × 8.85 × 10–12 (2.8 × 10–15)2] = 29 N (1)
use of r = 1.4 × 10–15 m (–1) once only                                                      2

  (ii)   FG = m2 G / r2 (1)
= (1.67 × 10–27)2 × 6.67 × 10–11 / (2.8 × 10–15)2 = 2.4 × 10–35 N (1)             2

(iii)    FS = 29 N / same as FE allow ecf (1)                                                          1

  (d)   FE >> FG so FG negligible / insignificant / can be ignored or AW (1)                    1

  (e)    (i)      FE = 0 (1)                                                                                                 1

(ii)     FG = 2.4 × 10–35 N (approx.) allow ecf (1)                                                  1

(iii)    FS = 2.4 × 10–35 N (approx.) (1)

         comment: FS now repulsive (not attractive) or AW
or indicated by minus sign with FS; (1) any 3                                             1

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3.       (i)      4.5 (N kg–1)                                                                                                       1

(ii)     g = (–)GM/r2                                                                                                     1

(iii)    g ∞ 1/r2;so value is 40/9 = 4.4(4) (N kg–1) ecf c(i)                                                2

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  4.     (a)     force (of attraction) on unit mass (at that point in space/at the
surface of a planet) (1)                                                                                       1

  (b)   (i)      (mgh =) 1500 × 40 × 1.5 × 105; = 9.0 × 109 (J)                                           2

(ii)     (larger as) g decreases with height                                                             1

(iii)    v = 2πR/T; = 2π × 2.0 × 107/(4.5 × 103) = 2.8 × 104 m s–1 (2)
½ mv2; = 0.5 × 1500 × (2.8 × 104)2 = 5.9 × 1011 (J) (2)                               4
aliter: F = mv2/R; = mg; so ½ mv2 = ½ mgR;= 6.0 × 1011 (J)

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  5.     (a)     arrow towards centre of planet                                                                            1

(b)     (i)      g = GM/R2                                                                                               1

(ii)     gS/gO = R2/25R2; gS = 40/25 (= 1.6 N kg–1)                                                2

(iii)    gC/gO = R2/16R2 giving gS = 40/16 (= 2.5 N kg–1)                                       1

(iv)    average g = (2.5 + 1.6)/2 = 2.(05) (1)
Δp.e. (= mgavR) = 3.0 × 103 × 2.05 × 2.0 × 107; = 1.2 × 1011 (J) (2)             3

  (c)    g = v2/r; = 4π2(5R)/T2 (2)
1.6 = 4 × 9.87 × 1.0 × 108/T2 giving T2 = 24.7 × 108 and T = 5.0 × 104 (s) (2)        4

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  6.     (a)     i. F = GMm/r2 or F α Mm/r2 with labels (1)                                                         1
ii. finite universe contracts/ resultant force on stars (1)                                         1

  (b)   Any 2 from
i. (satellite B) has larger circumference/smaller velocity
(satellite B) Gravitational field strength is less
(satellite B) Centripetal force is less                                                                    2

          ii.(accept calculation from either satellite)
r13/ T12 = r23/ T22 (1)
satellite A                                satellite B
r23 = 70003 × 57.22 / 1.632        r23 = 671003 × 57.22 / 1.632 (1)
r2 = 75,030 km                        r2 = 75, 320 km (1)                                             3
(= 75,000 km)                          (= 75,000 km)

(c)     Land-based are (any 3) 1 mark for each
more light can be collected/ made larger
more stable
more manoeuvrable
cheaper to build/repair
longer lifetime/ not exposed to high velocity particles
greater access                                                                                                    3

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  7.     It is the force (of attraction) per unit mass.                                                                  B1

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  8.     (a)     (i)      Force per unit mass (placed at that point) (1)                                              1

(ii)     g = GM/R2 (1)                                                                                          1

(iii)    Choosing a correct pair of values from the graph, e.g. 6.4 × 106 & 9.8,
10 × 106 & 4.0; (1)
substitute, 9.8 = 6.67 × 10–11 × M/ (6.4 × 106)2 to show
M = 6.0 × 1024 kg (1)                                                                                2

  (iv)  linear graph through origin/from 0 to R (1)                                                 1

  (v)   64 km; 1/100 of R as linear graph under Earth/AW (2)
64000 km; g  1/r2 so for 1/100 g r = 10R (2)                                           4

  (b)   GMe/R12 = GMm/R22; Me = 81 Mm; (2)
Mm = 6.0 × 1024/81 = 7.4 × 1022 (kg) ecf a(iii) (1)                                                3
or any acceptable alternative method i.e. correct method; correct figures;
processed to correct answer

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  9.     (i)      The gravitational field strength g is not constant.                                               B1
The student’s value would be greater than the actual value (because the
average magnitude of g is less than 9.81 m s–1).                                                 B1

  (ii)   KE = 1/2 mv2
v = 2πr / T                                                                                                       C1
v = 2 × π × (6800 + 6400) × 103 / 8.5 × 103 / v = 9.76 × 103 (m s–1)                    C1
KE = 1/2 × 1500 × (9.76 × 103)2
KE = 7.1(4) × 1010 (J)                                                                                     A1

  (iii)  A geostationary satellite stays above the same point on the Earth and as             B1
such can be used for radio communications. (the term communications to          M1
be included and spelled correctly to gain the mark).                                           A1
The satellite is not in geostationary orbit
because its period is less than 1 day / 8.6 × 104 s.

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  10.   (i)      r has been increased by a factor of 3 from the centre of planet.                           C1
g = (40/32 =) 4.4(4) (N kg–1)                                                                            A1

  (ii)   M = gr2 / G
M = (40 × [2.0 × 107]2) / 6.67 × 10–11                                                               C1
M = 2.4 × 1026 (kg)                                                                                         A1

(iii)    M = ρV = 4/3 πr3 ρ                                                                                          M1
g = GM / r2  r3 / r2 (Hence g  r)                                                                 A1

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  11.   The astronaut is accelerating / has centripetal acceleration (1)
and the space station has the same acceleration (1)
a person does not feel gravity (1)
only feels forces applied by contact with the walls of the space station (1)
no support force from the space station (as they have the same acceleration) (1)              4

MAXIMUM (4)

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