When a spacecraft travels away from Earth, the gravitational force acting on it

2 answers:

6 0

The correct answer is decreases

The further away you are the weaker it would be. That's why at one point you stop being in the field and ti doesn't pull you towards it anymore. Proportionally, if you move towards the Earth then it increases.

Sindrei [870]3 years ago

4 0

Gravitational force on aircraft at some distance from the Earth is given as

$F = \frac{Gm_1m_2}{r^2}$

here we know that

$m_1$ = mass of Earth

$m_2$ = mass of aircraft

r = distance between center of earth and aircraft

now as the aircraft is flying away from the surface of Earth so here the distance between them is increasing with the motion of aircraft

So as it moves away the force between them will decrease

So here we can say that

When a spacecraft travels away from Earth, the gravitational force acting on it

<u>decreases</u>

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A force of 5N and a force of 8N act to the same point and are inclined at 45degree to each other. Find the magnitude and directi

Alex_Xolod [135]

Magnitude: 12.1 N.
Direction: 17.0° to the 8 N force.

<h3>Explanation</h3>

Refer to the diagram attached (created with GeoGebra). Consider the 5 N force in two directions: parallel to the 8 N force and normal to the 8 N force.

$\displaystyle F_{\text{1, Parallel}} = F_1 \cdot \cos{45^\textdegree} = \dfrac{5\sqrt{2}}{2}\;\text{N}$ .
$\displaystyle F_{\text{1, Normal}} = F_1 \cdot \sin{45^\textdegree} = \dfrac{5\sqrt{2}}{2}\;\text{N}$ .

The sum of forces on each direction will be the resultant force on that direction:

Resultant force parallel to the 8 N force: $(8 + \dfrac{5\sqrt{2}}{2})\;\text{N}$ .
Resultant force normal to the 8 N force: $\dfrac{5\sqrt{2}}{2}\;\text{N}$ .