Back
to top
Gravitation field strength help
Watch
Announcements
Page 1 of 1
Go to first unread
Skip to page:
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?
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
reply
Report
#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
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
reply
(Original post by Adam_1999)
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
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
And you misread it, says 10^3km so would be 10^6m.
It's the next part I'm confused with.
0
reply
Report
#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.
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.
0
reply
(Original post by Adam_1999)
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.
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.
0
reply
Report
#6
(Original post by Super199)
How did you go about getting that equation
How did you go about getting that equation
g1 proportional to 1/r1²
g2 proportional to 1/r2²
Now find g1/g2
1
reply
(Original post by metrize)
From proportionality
g1 proportional to 1/r1²
g2 proportional to 1/r2²
Now find g1/g2
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
reply
Report
#8
(Original post by Super199)
Right ok what about this one.
Why can't you use g= GM/r^2
and just divide by the stuff it says. Part bi
![Name: aa.jpg
Views: 56
Size: 39.1 KB]()
Right ok what about this one.
Why can't you use g= GM/r^2
and just divide by the stuff it says. Part bi
You get the mass of earth and radius in your formula sheet right?
0
reply
(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?
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?
How would you use proportionalities in this case?
If mass is a factor?
0
reply
Report
#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?
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 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
0
reply
(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
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
0
reply
Report
#12
(Original post by Super199)
Interesting, though I always get confused on what things go in the proportionality (variables)
Interesting, though I always get confused on what things go in the proportionality (variables)
0
reply
X
Page 1 of 1
Go to first unread
Skip to page:
Quick Reply
Today on TSR
-
Ofqual consultationTeachers to use 'mini-exams' and other work to decide grades
- Is it too late to do well in my A-levels?
- What Harry Potter house are you in?
- 2021 wellbeing blog launch
- Student tips on lockdown mental health