An image illustrating the concept of physics and astronautics. The picture should depict a scene within a spacecraft, where a Caucasian male astronaut is preparing for the launch. The astronaut is situated inside the spacecraft, surrounded by various control panels and dials. Outside the window of the spacecraft, the view of the Earth below should be partially visible. The scene unveils the anticipation of the astronaut feeling the gravitational pull which changes during the vertical launch. The image should not contain any text.

An astronaut of mass m in a spacecraft experiences a gravitational force F=mg when stationary on the launchpad.

What is the gravitational force on the astronaut when the spacecraft is launched vertically upwards with an acceleration of 0.2g?

A 1.2mg
B mg
C 0.8mg
D 0

Ans= B
Someone explain pleaseeee!
And working too!
Thanks!

Gravitational force does not alter when the spacecraft is launched. So it is B

Well, let me explain it to you in my own unique way!

When the astronaut is stationary on the launchpad, they experience a gravitational force of F = mg, where m is the mass of the astronaut and g is the acceleration due to gravity.

Now, when the spacecraft is launched vertically upwards with an acceleration of 0.2g, the astronaut will still experience the same gravitational force, which is mg. The additional acceleration of the spacecraft does not affect the gravitational force acting on the astronaut.

So, the gravitational force on the astronaut when the spacecraft is launched vertically upwards with an acceleration of 0.2g is still mg. In other words, the gravitational force remains the same as when the astronaut was on the launchpad, so the answer is B - mg.

I hope that explanation brightened up your day a little bit!

To find the gravitational force on the astronaut when the spacecraft is launched vertically upwards with an acceleration of 0.2g, we can consider the forces acting on the astronaut.

When the astronaut is stationary on the launchpad, the gravitational force they experience is F = mg, where m is the mass of the astronaut and g is the acceleration due to gravity.

When the spacecraft is launched vertically upwards with an acceleration of 0.2g, there are two forces acting on the astronaut: the gravitational force and the apparent weight of the astronaut due to the acceleration.

The apparent weight of the astronaut due to the acceleration can be calculated using Newton's second law, which states that the net force on an object is equal to the mass of the object multiplied by its acceleration. In this case, the net force is the apparent weight of the astronaut, the mass is m, and the acceleration is 0.2g. Therefore, the apparent weight due to the acceleration is F_apparent = m * 0.2g.

The total force on the astronaut is the sum of the gravitational force and the apparent weight due to the acceleration. Therefore, the total force is F_total = F + F_apparent = mg + m * 0.2g.

We can simplify this equation by factoring out the common factor of m: F_total = m(g + 0.2g).

Now we can find the gravitational force on the astronaut when the spacecraft is launched vertically upwards with an acceleration of 0.2g by substituting F_total = mg + m * 0.2g into the given options:

A) 1.2mg: This is not the correct answer because it is greater than the sum of the gravitational force and the apparent weight due to the acceleration.

B) mg: This is the correct answer. The gravitational force alone is mg, without considering the acceleration.

C) 0.8mg: This is not the correct answer because it is less than the sum of the gravitational force and the apparent weight due to the acceleration.

D) 0: This is not the correct answer because the astronaut still experiences the gravitational force even when the spacecraft is launched vertically upwards.

Therefore, the correct answer is B) mg, which represents the gravitational force on the astronaut when the spacecraft is launched vertically upwards with an acceleration of 0.2g.

The gravitational force is m g

However the spring scale that astronaut is standing on would show 1.2 m g because the total force up = m g + ma
because
force up from scale - m g down = m a

To solve this problem, we need to understand the concept of apparent weight.

When the spacecraft is stationary on the launchpad, the astronaut experiences a gravitational force of F = mg, where g is the acceleration due to gravity.

When the spacecraft is launched vertically upwards with an acceleration of 0.2g, the astronaut also experiences an additional force due to this acceleration. This force is what we call the apparent weight.

The apparent weight is given by the formula:

Apparent weight = Actual weight + Additional force

In this case, the actual weight of the astronaut is mg, and the additional force is equal to ma, where a is the acceleration of the spacecraft.

So, the apparent weight is:

Apparent weight = mg + ma

Substituting the given values, with a being 0.2g, we get:

Apparent weight = mg + (0.2g)(m)

Simplifying, we get:

Apparent weight = mg + 0.2mg

Combining like terms, we have:

Apparent weight = (1 + 0.2)mg

Simplifying further, we get:

Apparent weight = 1.2mg

Therefore, the gravitational force on the astronaut when the spacecraft is launched vertically upwards with an acceleration of 0.2g is 1.2 times the astronaut's actual weight, which is option A.

The answer is B because, in reality, the force of gravity on astronauts, just depends on their masses which do not change, though the spacecraft is subjected to an acceleration upward.

Due to the acceleration of the spacecraft, is exerted on the astronaut additional acceleration equal to 0.2 g. In this case the astronaut has a greater apparent weight, which is equal to 1.2 mg although the gravitational force of the earth on the mass of the astronaut is still mg. Let's see this with the second law of Newton.
By Newton's second law we have:
∑ F = ma
The forces acting on an astronaut are the force of gravity or the weight, directed downward, and the force of the rocket boosters Fr, directed upwards. Then we can write:
Fr – mg = ma
Where (a = 0.2g) is the acceleration of the spacecraft ; so we have:
Fr = ma + mg → Fr = m(a + g) → Fr = m(0.2g + g) → Fr = 1.2mg
Fr is the apparent gravitational force (apparent weight) experienced by astronauts