Friday

December 19, 2014

December 19, 2014

Posted by **ikye** on Friday, July 5, 2013 at 10:59am.

- phy -
**Steve**, Friday, July 5, 2013 at 11:06amQ1

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 (√2-1)R

(√2-1) * 64000000 = 26500000

so I'd choose (a)

Use similar logic on the others.

- phy -
**Elena**, Friday, July 5, 2013 at 1:15pmQ1

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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