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Home Q & A Board Science-Related Homework Help Physics
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buggingout16 buggingout16
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11 years ago
A hypothetical planet has a radius 2.1 times that of Earth, but has the same mass.

What is the acceleration due to gravity near its surface?
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wrote...
#1
11 years ago
Gravitational attraction is inversely proportional to the square of the distance (or radius) between the centers of two bodies, and do other factor has changed.
The acceleration on Earth is 9.8m/s^2 and so on the hypothetical planet with 2.1 times the radius of earth is will be 1/(2.1)^2 times the acceleration on Earth, so it is 2.22m/s^2
wrote...
#2
11 years ago
Acelleration due to gravity is 9.8 m/s
RJ
wrote...
#3
11 years ago
it depends on how large the obgect is the lager the mass the higher the terminal velosity
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#4
11 years ago
according to this:
g=GM/(R+h)^2    (g=gravity, G=gravity constant, M=mass, R=radius, h= height)

x=planet
g/gx=
[GM/(R+h)^2]/[GMx/(Rx+hx)^2]
because u said "due to gravity near its surface":
h=hx=0
and also we have Mx=M and Rx=2.1R
So:
g/gx=(1/R^2)/(1/(2.1R)^2);
g/gx=(2.1R)^2/(R^2);
g/gx=4.41;
gx=(1/4.41)g
wrote...
#5
11 years ago
no no no no no im in AP Physics i can help u... ha ah
ok

g=Gm/d^2

there for when u use the mass of the earth for m and u use the constant G for G and u use 2.1 times the radius of the earth u get the answer

3.987*10^14/(6378137*2.1)^2
3.987*10^14/1.794*10^14

finalllllyyy u get 2.22 m/s/s
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