Respuesta :
Answer:
Explanation:
Let mass of the apple be m . Force on it at surface of the planet can be expressed as follows
F_R = G Mm / R²
Where M is mass of the planet.
Let at at height h the gravitational force on apple be 0.3 F_R
0.3 F_R = G Mm / ( R +h)²
Dividing the two equation
1 / 0.3 = ( R +h)²/ R²
( R +h)/ R = 1.825
h = .825 R
b )
If we go deep inside the planet to get force on apple equal to .3 F_R
.3 F_R = G 4/3 π ( R - h )³ d m / (R-h)² [ d is density of planet ]
.3 F_R = G 4/3 π ( R - h ) d m
F_R = G Mm / R²
F_R = G 4/3πR³d m / R²
F_R = G 4/3πR d m
dividing two equations
0.3 = ( R - h ) / R
h = .7 R
The answers to your question are as listed below
A) If we move the apple 0.3 FR away from the planet the value of h is ;
0.825 R
B) If we move the apple 0.3 FR into the tunnel the distance is ; 0.7 R
Assumptions :
mass of apple = m
Force on apple at planet surface = [tex]F_{r}[/tex] = GMm / R² --- ( 1 )
Determine th value of h
a) Applying equation ( 1 )
gravitational force of apple at h = 0.3 FR
therefore equation ( 1 ) becomes
0.3 FR = GMm / (R + h)² ----- ( 2 )
divide both sides of equation by
1 / 0.3 = ( R + h )²/ R²
therefore :
h = 0.825 R
b) When we go into the tunnel
Applying the relation below
0.3 FR = G *4/3 π *( R - h )³* dm / (R-h)² -- ( 3 )
where : d = density of planet
resolving equation ( 3 )
FR = G *4/3πR * dm
dividing both sides of the equation
0.3 = ( R - h ) / R
therefore :
h = 0.7 R
Hence we can conclude that A) If we move the apple 0.3 FR away from the planet the value of h is 0.825 R and If we move the apple 0.3 FR into the tunnel the distance is ; 0.7 R
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