Posted by ikye on Friday, July 5, 2013 at 10:59am.
(Q1)At what altitude above the earth's surface would the acceleration due to gravity be 4.9m/s2?Assuming the mean radius of the earth is 64000000meters and the acceleration due to gravity 9.8m/s2 on the surface of the earth (a)26000000m(b)323000000m(c)465000000m(d)776000000m(Q2)A man weighs 750 N on the surface of the earth.What will be his weight when standing on the moon? The masses of the earth and the moon are respectively 59800000000kg and 73600000 kg.Their radii are respectvely 637000km and 174000km (a)200.5N(b)123.7(c)550.4N(d)1000.0N.(Q3)A 2000kg satelite orbits the earth at a height of 300km. What is the speed of the satellite and its period?Take G=0.0000000000667Nm2/kg2 Mass of earth is 598000000kg (a)7.73 km/s 54000(b)855.4km/s and 770000(c)497.2km/s 5500000(d)322.3km/s and 430000

phy  Steve, Friday, July 5, 2013 at 11:06am
Q1
You want to have 1/2 the gravitational acceleration at the surface.
Since F ∝ 1/r^2, you will need the new height to be √2 times the surface radius. That is, at an altitude of (√21)R
(√21) * 64000000 = 26500000
so I'd choose (a)
Use similar logic on the others.

phy  Elena, Friday, July 5, 2013 at 1:15pm
Q1
the gravitational constant G =6.67•10⁻¹¹ N•m²/kg²,
Radius of the Earth is 6400 000 m =6.4•10⁶ m ===Your data is incorrect!!!
mg=G•m•M/R²
mg`= G•m•M/(R+h)²
(R+h)²/ R² = g /g`=9.8/4.9 = 2
R²+2Rh + h²= 2R²
h²+2R  R²= 0
h =  R ±sqrt(2R²) =  R ± R√2 =>
h=R(√2 1) = 0.41 R=
=0.41 •6.4•10⁶ =2624000 m
Ans. (a)
Q2
Mass of the Earth = 5.98•10²⁴ kg,
Radius of the Earth = 6.4•10⁶ m,
Mass of the Moon = 7.36•10² kg,
Radius of the Moon = 1.74•10⁶ m,
mg= G•m•M/R²,
mg`= G•m•M`/R`²,
g/g`= M R`²/R² M`,
g`=g R² M`/ M R`²=
=9.8•7.36•10²²(6.4•10⁶)²/5.98•10²⁴(1.74•10⁶)² = =1.63.
W=mg => m =W/g=750/9.8 =76.53 kg
mg`= 76.53•1.63 = 124 N
Q3
mv²/(R+h) = G•m•M/(R+h)²,
v=sqrt{GM/(R+h)} =
=sqrt{6.67•10⁻¹¹•5.98•10²⁴/(6.4•10⁶ +3•10⁵) =
= 7730 m/s =7.73 km/s.
T=2π(R+h)/v =
= 2π(6.4•10⁶ +3•10⁵)/4730 ≈ 5400 s.
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