Gravity & Mass: Woman's Weight at 2x Earth's Radius

[babyjachy]

Homework Statement

a woman has a mass of 55 kg on Earth's surface. what is her weight at a height at twice the Earth's radius? the Earth's radius is 6.4 x 10 ^ 6 meters..

Homework Equations

w = mg

The Attempt at a Solution

of course, the weight of the woman at that particular height will decrease..but i can't get the correct answer.

please help :)

 

[gneill]

Hi babyjachy, welcome to Physics Forums.

What have you tried? Show some work and then we can help.

 

[babyjachy]

this is my solution
(just tell me if I'm doing the right thing..

w = mxg
= (55kg)(9.8m/s^2/1.28x10^7)
= 4.21x10^-5
is it correct?

 

[Dick]

this is my solution
(just tell me if I'm doing the right thing..

w = mxg
= (55kg)(9.8m/s^2/1.28x10^7)
= 4.21x10^-5
is it correct?

You aren't using w=mg, you are using something else. What is it? And no, it's not even close to correct. Use Newton's law of Universal Gravitation. F=G*m1*m2/r^2. To start with, what's the weight at the Earth's surface?

 

[babyjachy]

you aren't using w=mg, you are using something else. What is it? And no, it's not even close to correct. Use Newton's law of universal gravitation. F=g*m1*m2/r^2. To start with, what's the weight at the Earth's surface?

the weight at the Earth's surface is 539 Newtons.

 

[Dick]

539 n..

Great! Now use F=G*m1*m2/r^2. That's the same F as F=mg except F=mg only works at the Earth's surface. The only thing that changes in the problem is that the radius is doubled. How does changing r to 2r affect F?

 

[babyjachy]

Great! Now use F=G*m1*m2/r^2. That's the same F as F=mg except F=mg only works at the Earth's surface. The only thing that changes in the problem is that the radius is doubled. How does changing r to 2r affect F?

F becomes less..thanks dick..god bless

 

[Dick]

F becomes less..thanks dick..god bless

Bless you too. But how much less? Did you get the whole problem? What is the weight at 2r?

 

[gneill]

Also, when the problem says "at a height at twice the Earth's radius", do they mean that the radial distance is 2R, or do they mean the height above the Earth's surface is twice the Earths radius? Wording in problems is sometimes ambiguous...

 

[Dick]

Also, when the problem says "at a height at twice the Earth's radius", do they mean that the radial distance is 2R, or do they mean the height above the Earth's surface is twice the Earths radius? Wording in problems is sometimes ambiguous...

Good eye, gneill. It's possible they mean 3R for the radius. Thanks for noticing. I didn't.

 

[babyjachy]

Good eye, gneill. It's possible they mean 3R for the radius. Thanks for noticing. I didn't.

thanks gneill and dick..i'll review my solution..
i think the problem means that the height is twice the radius of the earth..
so, will i use 3R instead?

 

[Dick]

thanks gneill and dick..i'll review my solution..
i think the problem means that the height is twice the radius of the earth..
so, will i use 3R instead?

I think it probably does. 'height' is usually distance from the Earth's surface, not from it's center, which is what I was thinking.

 

[babyjachy]

by the way, just want to check if my solution is correct..using the formula F = g*m1*m2/3r^2, will i substitute 55kg for m1 and m2?

 

[babyjachy]

by the way, just want to check if my solution is correct..using the formula F = g*m1*m2/3r^2, will i substitute 55kg for m1 and m2?

if that so, force of gravity is 4.913x10^-21..
am i on the right track?

 

[Dick]

by the way, just want to check if my solution is correct..using the formula F = g*m1*m2/3r^2, will i substitute 55kg for m1 and m2?

No. m1 is the mass of the earth. m2 the is mass of the lady. Look up Newtons Law. You CAN use that formula, but you don't have to. You know F=539 Newtons at the surface of the earth. Only r changes when you move to a different radius. Just figure out by what RATIO F must change. And it's (3r)^2 in the denominator, not 3(r^2). There is a difference.

 

[babyjachy]

No. m1 is the mass of the earth. m2 the is mass of the lady. Look up Newtons Law. You CAN use that formula, but you don't have to. You know F=539 Newtons at the surface of the earth. Only r changes when you move to a different radius. Just figure out by what RATIO F must change. And it's (3r)^2 in the denominator, not 3(r^2). There is a difference.

ok..what does it mean by "what ratio Force must change"?could you please elaborate.

 

[astro_physics]

No. m1 is the mass of the earth. m2 the is mass of the lady. Look up Newtons Law. You CAN use that formula, but you don't have to. You know F=539 Newtons at the surface of the earth. Only r changes when you move to a different radius. Just figure out by what RATIO F must change. And it's (3r)^2 in the denominator, not 3(r^2). There is a difference.

why r becomes (3r) ^2? I am confused..

 

[astro_physics]

i really find it hard to understand the problem. sigh..

 

[Dick]

Suppose F1=G*m1*m2/r^2 is the force at the Earth's surface which you know is 539 N. The force at 3r is F2=G*m1*m2/(3r)^2. What's F2/F1? Since you know F1, what is F2?

 

[Dick]

i really find it hard to understand the problem. sigh..

If 'astro_physics' and 'babyjachy' are the same person, you should be aware that having more than one identity in the Forums isn't allowed.

 

[Calrid]

ok..what does it mean by "what ratio Force must change"?could you please elaborate.

The force of gravity changes with distance by an inverse relation to a square of distance, which is why it is called the inverse squared law. You don't need to know why yet. You just need to know it does.

You have the equation now just plug in the numbers.

Gravitational force becomes weaker between two bodies as distance increases. What more is there to know in this question?

 

[gneill]

ok..what does it mean by "what ratio Force must change"?could you please elaborate.

It is often convenient to solve problems that have some number of unknowns that you know remain constant by way of setting up ratios. You probably remember doing this with the ideal gas law, PV = nRT, where you'd set up a ratio to find out how, say, temperature changes when the pressure changed and the volume remained constant. You didn't have to find the values of n, R, or V because they canceled out in the ratio.

In this case it's the gravitational law that you use. The law of gravitation is an inverse square law. That is, the force between two bodies varies as the inverse of the square of the distance between them. If the masses of the two bodies (m1 and m2) remain the same and only the distance changes, and everything else is constants, then it's a candidate for the ratio approach.

The force due to gravity on a body of mass m at radial distance r_o from the center of the Earth (which has mass M) is:

$$F = G\frac{M m}{r_o^2}$$

Divide both sides by m to find the acceleration due to gravity at radial distance r.

$$g = \frac{G M}{r_o^2}$$

At some other radial distance, say r₁, the acceleration due to gravity would be:

$$g_1 = \frac{G M}{r_1^2}$$

Note that G and M are constants. Set up the ratio g₁/g_o and see what happens.

 

[babyjachy]

If 'astro_physics' and 'babyjachy' are the same person, you should be aware that having more than one identity in the Forums isn't allowed.

astrophysics and babyjachy are two different people..we are friends and we both want to solve that particular problem..swear..

 

[Dick]

astrophysics and babyjachy are two different people..we are friends and we both want to solve that particular problem..swear..

Then no problem.

 

[babyjachy]

thanks dick and gneill..more power