1. What force does the earth's gravity exert on a 81kg astronaut who is orbiting in a space station at a distance from the SURFACE of the earth equal to 2 earth radii? answer must be in newtons.

2. The astronaut in the above was launched into space by a Saturn rocket. The upward acceleration of a Saturn rocket shortly after blast-off in 80m/s^2. When the Saturn rocket is accelerating, what is the apparent weight of astronaut, that is, what does the astronaut experience as the weight of his body?(Hint: Keep in mind the definition of weight and that more than acceleration due to gravity is acting on the astronaut.) Answer must be in Newtons.

TIA!

Both of these questions have us stumped.

# Physics help, please

Started by
1cat2ferrets
, Mar 05 2010 03:42 PM

4 replies to this topic

### #2

Posted 05 March 2010 - 04:01 PM

I'd post this on the high school board...you might get some physics pros there!

### #3

Posted 05 March 2010 - 09:32 PM

[quote name='1cat2ferrets']1. What force does the earth's gravity exert on a 81kg astronaut who is orbiting in a space station at a distance from the SURFACE of the earth equal to 2 earth radii? answer must be in newtons. [/QUOTE]

Problem 1

-----------

Away from earth we know that there is loss in weight. The question is asking to find how much will astronaut weigh in Newtons the space station up there.

So astronaut with mass 81 kg has weight (81 kg) x (10 m/s^2) = 810 N when on earth. So expect answer to be much less than 810 N.

By Newton's law of gravitation

F = (GMm)/d^2 = ma ----------> (1)

Where

G = 6.67 x 10^(-11) Nm^2/kg^2 = constant of gravitation

M = 5.98 x 10^24 kg = Mass of earth

m = 81 kg = astronaut's mass

R = 6.37 x 10^6 m =radius of earth

d = 3R = distance between center of earth and the astronaut

Substitute relevant variable in (1)

Force in Newton = [6.67 x 10^(-11)] * [5.98 x 10^24] * [81] / (3 * 6.37 x 10^6)^2

= 88.47 N ( which on earth is 810 N so we are doing ok)

2. The astronaut in the above was launched into space by a Saturn rocket. The upward acceleration of a Saturn rocket shortly after blast-off in 80m/s^2. When the Saturn rocket is accelerating, what is the apparent weight of astronaut, that is, what does the astronaut experience as the weight of his body?(Hint: Keep in mind the definition of weight and that more than acceleration due to gravity is acting on the astronaut.) Answer must be in Newtons.[/QUOTE]

Problem 2

------------

When astronaut is sitting in the seat while rocket is still on ground his seat is pushing him/her upward with 810 N force which is the weight. When rocket itself has acceleration in upward direction there will be additional force of (80 m/s^2) * (81 kg) = 6480 N

So, net upward force experienced by astronaut = 6480 + 810 = 7290 N

You can imagine astronaut's upward force on a moderate scale when you go upwards in a very high speed elevators in tall buildings.

[QUOTE]

TIA!

Both of these questions have us stumped.[/QUOTE]You are welcome.

Please check answers in your book if you can. I did these problems in a bit of hurry. If my solutions are incorrect, please post your feedback. Thanks.

Also, if you had posted the question under

http://www.welltrain...ad.php?t=153209

I would have received email from forum admin about your post and I would have known about your questions a little earlier.

Best regards.

MPCTutor

---------------------------------------------------------------------------

AP Calculus, AP Physics, Singapore Math Grades 7-12

---------------------------------------------------------------------------

US Central Time: 8:31 PM 3/5/2010

Problem 1

-----------

Away from earth we know that there is loss in weight. The question is asking to find how much will astronaut weigh in Newtons the space station up there.

So astronaut with mass 81 kg has weight (81 kg) x (10 m/s^2) = 810 N when on earth. So expect answer to be much less than 810 N.

By Newton's law of gravitation

F = (GMm)/d^2 = ma ----------> (1)

Where

G = 6.67 x 10^(-11) Nm^2/kg^2 = constant of gravitation

M = 5.98 x 10^24 kg = Mass of earth

m = 81 kg = astronaut's mass

R = 6.37 x 10^6 m =radius of earth

d = 3R = distance between center of earth and the astronaut

Substitute relevant variable in (1)

Force in Newton = [6.67 x 10^(-11)] * [5.98 x 10^24] * [81] / (3 * 6.37 x 10^6)^2

= 88.47 N ( which on earth is 810 N so we are doing ok)

2. The astronaut in the above was launched into space by a Saturn rocket. The upward acceleration of a Saturn rocket shortly after blast-off in 80m/s^2. When the Saturn rocket is accelerating, what is the apparent weight of astronaut, that is, what does the astronaut experience as the weight of his body?(Hint: Keep in mind the definition of weight and that more than acceleration due to gravity is acting on the astronaut.) Answer must be in Newtons.[/QUOTE]

Problem 2

------------

When astronaut is sitting in the seat while rocket is still on ground his seat is pushing him/her upward with 810 N force which is the weight. When rocket itself has acceleration in upward direction there will be additional force of (80 m/s^2) * (81 kg) = 6480 N

So, net upward force experienced by astronaut = 6480 + 810 = 7290 N

You can imagine astronaut's upward force on a moderate scale when you go upwards in a very high speed elevators in tall buildings.

[QUOTE]

TIA!

Both of these questions have us stumped.[/QUOTE]You are welcome.

Please check answers in your book if you can. I did these problems in a bit of hurry. If my solutions are incorrect, please post your feedback. Thanks.

Also, if you had posted the question under

http://www.welltrain...ad.php?t=153209

I would have received email from forum admin about your post and I would have known about your questions a little earlier.

Best regards.

MPCTutor

---------------------------------------------------------------------------

AP Calculus, AP Physics, Singapore Math Grades 7-12

---------------------------------------------------------------------------

US Central Time: 8:31 PM 3/5/2010

### #4

Posted 06 March 2010 - 12:43 PM

OK, I'll take a stab at this, too.

Prob #1:

I agree with the answer to Problem 1 above, but my method is a bit different:

The gravitational attraction between the two bodies (man and earth) is given by: (and the gravitational

force of the earth acting on the man is just his "weight")

F = (G*m1*m2/r^2) where m1 = mass of man, m2 = mass of earth, G = gravitational constant

and r = distance between man and the earth's center

When the man stands on the surface of the earth, r = re = radius of earth

When the man is in space (2 earth radii above the earth's surface), r = 3 * re

Let's take a ratio of the the force of gravity acting on the man while he's standing on the earth's surface

(let's call this Fe) to the force of gravity acting on the man while he's orbiting in space (let's call this Fo):

Fe/Fo = (G*m1*m2/re^2) / (G* m1*m2/ (3*re)^2)

= (3*re)^2/re^2 ...since G, m1, m2 are the same in the numerator and in the denominator

=(9* re^2)/re^2

=9

So the force of gravity exerted on the man is 9 times greater on the earth's surface than when he's in orbit.

Force exerted by gravity on orbiting man = (1/9) * force of gravity on man on earth's surface

= (1/9) * (mass of man) * (9.8 m/s^2)

= (1/9) * (81 kg) * (9.8 m/s^2)

= 9 * 9.8 N = 88.2 N = 88 N (significant figures)

Prob #2:

The apparent weight of the astronaut = mass of astronaut * net acceleration on astronaut

Now the net acceleration has two components, one due to the earth's gravity (9.8 m/s^2 directed downward)

and another due to the accelerating rocket (80 m/s^2) again directed downward. (mpcTutor explained this using elevators)

So apparent weight = (81 kg) * (9.8 m/s^2 + 80 m/s^2)

= (81 * 89.8) N

= 7273.8 N

or 7300 N (significant figures)

Prob #1:

I agree with the answer to Problem 1 above, but my method is a bit different:

The gravitational attraction between the two bodies (man and earth) is given by: (and the gravitational

force of the earth acting on the man is just his "weight")

F = (G*m1*m2/r^2) where m1 = mass of man, m2 = mass of earth, G = gravitational constant

and r = distance between man and the earth's center

When the man stands on the surface of the earth, r = re = radius of earth

When the man is in space (2 earth radii above the earth's surface), r = 3 * re

Let's take a ratio of the the force of gravity acting on the man while he's standing on the earth's surface

(let's call this Fe) to the force of gravity acting on the man while he's orbiting in space (let's call this Fo):

Fe/Fo = (G*m1*m2/re^2) / (G* m1*m2/ (3*re)^2)

= (3*re)^2/re^2 ...since G, m1, m2 are the same in the numerator and in the denominator

=(9* re^2)/re^2

=9

So the force of gravity exerted on the man is 9 times greater on the earth's surface than when he's in orbit.

Force exerted by gravity on orbiting man = (1/9) * force of gravity on man on earth's surface

= (1/9) * (mass of man) * (9.8 m/s^2)

= (1/9) * (81 kg) * (9.8 m/s^2)

= 9 * 9.8 N = 88.2 N = 88 N (significant figures)

Prob #2:

The apparent weight of the astronaut = mass of astronaut * net acceleration on astronaut

Now the net acceleration has two components, one due to the earth's gravity (9.8 m/s^2 directed downward)

and another due to the accelerating rocket (80 m/s^2) again directed downward. (mpcTutor explained this using elevators)

So apparent weight = (81 kg) * (9.8 m/s^2 + 80 m/s^2)

= (81 * 89.8) N

= 7273.8 N

or 7300 N (significant figures)

**Edited by Kathy in Richmond, 06 March 2010 - 07:22 PM.**

### #5

Posted 06 March 2010 - 03:34 PM

for the answers. We needed the explanations and both helped quite a bit.