# Gravitation field strength help

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#1
part ii)

I tried using g= GM/R^2

But how come you can't use 6.67*10^-11 and have to use 3.7 for G?
0
3 years ago
#2
g = GM/r^2
g is the gravitational field strength (N kg^-1)
G is a constant
M is the mass of the object CREATING the field (kg)
r is the distance between the centre of mass of the object and the point in gravitational field you are interested in (m)

3.7 = (6.67 x 10^-11 x M)/(3.4 x 10^3)^2

Solving for M gives M = 6.41 x 10^17 kg
0
#3
g = GM/r^2
g is the gravitational field strength (N kg^-1)
G is a constant
M is the mass of the object CREATING the field (kg)
r is the distance between the centre of mass of the object and the point in gravitational field you are interested in (m)

3.7 = (6.67 x 10^-11 x M)/(3.4 x 10^3)^2

Solving for M gives M = 6.41 x 10^17 kg
Thats part i)

And you misread it, says 10^3km so would be 10^6m.

It's the next part I'm confused with.
0
3 years ago
#4
g1/g2 = (r2/r1)^2.

Set g1 as field strength at the surface, and g2 as the field strength you want to find. Rearrange the above equation and sub in the values to find g2.
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#5
g1/g2 = (r2/r1)^2.

Set g1 as field strength at the surface, and g2 as the field strength you want to find. Rearrange the above equation and sub in the values to find g2.
How did you go about getting that equation
0
3 years ago
#6
(Original post by Super199)
How did you go about getting that equation
From proportionality

g1 proportional to 1/r1²
g2 proportional to 1/r2²

Now find g1/g2
1
#7
(Original post by metrize)
From proportionality

g1 proportional to 1/r1²
g2 proportional to 1/r2²

Now find g1/g2
Why can't you use g= GM/r^2

and just divide by the stuff it says. Part bi

0
3 years ago
#8
(Original post by Super199)
Why can't you use g= GM/r^2

and just divide by the stuff it says. Part bi

Yeah you could, normally for proportionaluty stuff I like doing the ratio method since it simplifies it. But both is fine.

You get the mass of earth and radius in your formula sheet right?
0
#9
(Original post by metrize)
Yeah you could, normally for proportionaluty stuff I like doing the ratio method since it simplifies it. But both is fine.

You get the mass of earth and radius in your formula sheet right?
Im not sure, i tried it using the first method but didnt get the right answer.

How would you use proportionalities in this case?

If mass is a factor?
0
3 years ago
#10
(Original post by Super199)
Im not sure, i tried it using the first method but didnt get the right answer.

How would you use proportionalities in this case?

If mass is a factor?
Let g(e) be proportional to m(e)/(r(e))²
Let gm be proportional to m(m)/(r(m))²

So g(m)/g(e) = m(m)(r(e))²/(r(m))²m(e)

Then replace m(e) with 81m(m) and r(e) with 3.7r(m)

Then you can cancel stuff and multiple by g(e) leaving something times g on earth and finally replace g of earth with 9.81

Sorry if it's hard to look at formatting isn't my strong point
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#11
(Original post by metrize)
Let g(e) be proportional to m(e)/(r(e))²
Let gm be proportional to m(m)/(r(m))²

So g(m)/g(e) = m(m)(r(e))²/(r(m))²m(e)

Then replace m(e) with 81m(m) and r(e) with 3.7r(m)

Then you can cancel stuff and multiple by g(e) leaving something times g on earth and finally replace g of earth with 9.81

Sorry if it's hard to look at formatting isn't my strong point
Interesting, though I always get confused on what things go in the proportionality (variables)
0
3 years ago
#12
(Original post by Super199)
Interesting, though I always get confused on what things go in the proportionality (variables)
It's just the stuff you get given in the question pretty much, as everything else you can assume is constant
0
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