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madreJ [45]
3 years ago
12

One of the asteroids, Ida, looks like an elongated potato. Surprisingly it has a tiny (compared to Ida) spherical moon! This moo

n called Dactyl has a mass of 4.20 × 10^16 kg, and a radius of 1.57 × 10^4 meters, according to Wikipedia. Ida has a radius of 3.14 x 10^4 meters.
Find the acceleration of gravity on the surface of this little moon.
Physics
1 answer:
Mashcka [7]3 years ago
8 0

Answer:

g = 0.0114 m/s²

Explanation:

The value of acceleration due to gravity on the surface of the moon can be given by the following formula:

g = \frac{Gm}{r^2}

where,

g = acceleration due to gravity on the surface of moon = ?

G = Universal Gravitational Constant = 6.67 x 10⁻¹¹ N.m²/kg²

m = mass of moon = 4.2 x 10¹⁶ kg

r = radius of moon = 1.57 x 10⁴ m

Therefore,

g= \frac{(6.67\ x\ 10^{-11}\ N.m^2/kg^2)(4.2\ x\ 10^{16}\ kg)}{(1.57\ x\ 10^4\ m)^2}

<u>g = 0.0114 m/s²</u>

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<em>There is not enough data to solve the problem, but I'm assuming the initial height as h = 10 m for the question to have a valid answer and the student can have a reference to solve their own problem</em>

Answer:

(a) \Delta P=1.67 \ kg.m/s

(b) \Delta P=0.86\ m/s

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The momentum of a given particle of mass m traveling at a speed v is given by

P=m.v

When this particle changes its speed to a value v', the new momentum is

P'=m.v'

The change of momentum is

\Delta P=m.v'-m.v

\Delta P=m.(v'-v)

Defining upward as the positive direction, we'll compute the change of momentum in two separate cases.

(a) The initial height of the superball of m=61.6 gr = 0.0616 Kg is set to h= 10 m. This information leads us to have the initial potential energy of the ball just after it's dropped to the floor:

U=m.g.h=0.0616\cdot 9.8\cdot 10 =6.0368\ J

This potential energy is transformed into kinetic energy just before the collision occurs, thus

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Solving for v

\displaystyle v=\sqrt{\frac{6.0368\cdot 2}{0.0616}}

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h'=88.5\% (10)=8.85\ m

The potential energy reached by the ball at its rebound is

U'=m.g.h'=0.0616\cdot 9.8\cdot 8.85 =5.342568\ J

Thus the speed after the collision is

\displaystyle v'=\sqrt{\frac{5.342568\cdot 2}{0.0616}}

v'=13.17\ m/s

The change of momentum is

\Delta P=0.0616\cdot (13.17+14)

\Delta P=1.67 \ kg.m/s

(b) If the putty sticks to the floor, then v'=0

\Delta P=0.0616\cdot (0+14)

\Delta P=0.86\ m/s

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