If the 50 kg boy were in a spacecraft 5.0r from the center of the earth, what would his weight be?

1 answer:

7 0

·The acceleration of gravity is proportional to

   1 / (the square of the distance from the center) .

When we're on the surface, we're 1 radius from the center of the Earth,
and the acceleration of gravity is 9.8 m/s² .

The boy's weight = (mass) · (gravity) = (50kg) · (9.8 m/s²)

   = 490 newtons .

At the distance of 5 radii from the center (4 radii altitude from the surface),
the acceleration of gravity is

   (9.8 m/s²) · (1/5)² = 0.39 m/s² .

The boy's weight is (mass) · (gravity) = (50kg) · (0.39 m/s²)

= 19.6 newtons .

Just as we expected, his weight at that distance is

   (19.6 / 490) = 0.04 = 1/25 = 1/5² of his weight on the surface.

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all objects near the earths surface-regardless of size and weight have the same force of gravity acting on them. is it true or f

pickupchik [31]

False: the force of gravity acting on different objects is different and depends on their mass

Explanation:

The answer is false.

The force of gravity acting on an object (also known as weight) near the Earth's surface is given by:

$F=mg$

where:

m is the mass of the object

$g=9.8 m/s^2$ is the acceleration of gravity

We see from the formula that the force of gravity acting on an object depends on the mass: the larger the mass of the object, the stronger the gravitational force acting on it, and the smaller the mass, the weaker the force of gravity.

The factor that does not change is the acceleration of gravity, which is constant ( $9.8 m/s^2$ ) if we are near the Earth's surface, and implies that all the objects in free fall accelerate at the same rate towards the ground, regardless of their size and weight.