What is the weight of an astronaut before lift off in kilograms?

[Destiny153]

Homework Statement

so in my physics class, we are learning about the 3 laws of motion and all that jazz bla bla bla, i get all that stuff, just not the stuff where it comes to math. i suck at math! so we had homework, no duh, and well I am stuck, can i please have some help! anyone??

hear is the problem:

An astronaut has a mass of 50 kg.

A.) how much does she weigh before lift off?
B.) when her space vehicle is 6400 Km. above the Earth's suface, she will weigh one quarter of what she weighed on earth. what does she weigh at this point?

Homework Equations

so there you are, ps i know this formula but i don't understand it. Fnet = ma

net force= mass times acceleration? righ?

The Attempt at a Solution

for (A.) i think that it is 50kg, because Weight is proportional to mass; equal masses at the same location in a gravitational field will have equal weights

 

[SammyS]

for (A.) i think that it is 50kg, because Weight is proportional to mass; equal masses at the same location in a gravitational field will have equal weights

Proportional is not equal.

What's the definition of weight you learned in your course?

 

[Destiny153]

Weight: the force of gravity on the object and may be defined as the mass times the acceleration of gravity, w = mg

 

[SammyS]

So, what's the answer to (A) ? Kilograms is a unit of mass, not weight.

 

[rwisz]

. So, an astronaut's weight before lift off would be equal to their mass, which in this case is 50 kg.

For (B.), we need to use the formula W=mg, where W is weight, m is mass, and g is the acceleration due to gravity. We know that on Earth, the acceleration due to gravity is 9.8 m/s^2. So, on Earth, the astronaut's weight would be W=50kg * 9.8 m/s^2 = 490 N.

When the astronaut is 6400 km above the Earth's surface, the acceleration due to gravity decreases to approximately 3.2 m/s^2. So, using the same formula, the astronaut's weight at this point would be W=50kg * 3.2 m/s^2 = 160 N. This is one quarter of the weight on Earth, as stated in the problem.