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Novosadov [1.4K]
3 years ago
7

Suppose astronomers find an earthlike planet that is twice the size of Earth (that is, its radius is twice the radius of Earth).

What must be the mass of this planet such that the gravitational force (Fgravity) at the surface would be identical to Earth's?
Physics
1 answer:
JulijaS [17]3 years ago
6 0

Answer:

4 times the mass of Earth

Explanation:

M_1 = Mass of Earth

M_2 = Mass of the other planet

r = Radius of Earth

2r = Radius of the other planet

m = Mass of object

The force of gravity on an object on Earth is

F=\frac{GM_1m}{r^2}

The force of gravity on an object on the other planet is

F=\frac{GM_2m}{(2r)^2}

As the forces are equal

\frac{GM_1m}{r^2}=\frac{GM_2m}{(2r)^2}\\\Rightarrow M_1=\frac{M_2}{4}\\\Rightarrow M_2=4M_1

So, the other planet would have 4 times the mass of Earth

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A. The bomb will take <em>17.5 seconds </em>to hit the ground

B. The bomb will land <em>12040 meters </em>on the ground ahead from where they released it

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The equation for the height y with respect to ground in a horizontal movement (no friction) is

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With g equal to the acceleration of gravity of our planet and t the time measured with respect to the moment the bomb was released

The height will be zero when the bomb lands on ground, so if we set y=0 we can find the flight time

The range (horizontal displacement) of the bomb x is

x = v.t     [2]

Since the bomb won't have any friction, its horizontal component of the speed won't change. We need to find t from the equation [1] and replace it in equation [2]:

Setting y=0 and isolating t we get

t=\sqrt{\frac{2y_0}{g}}

Since we have y_0=1500m

t=\sqrt{\frac{2(1500)}{9.8}}

t=17.5 sec

Replacing in [2]

x = 688\ m/sec \ (17.5sec)

x = 12040\ m

A. The bomb will take 17.5 seconds to hit the ground

B. The bomb will land 12040 meters on the ground ahead from where they released it

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