Consider a neutron star v.i th a mass M equal to the mass of the sun, 1. 98 x 1030 kg, and a radius R of 12 km. \Vhat is the free-fall acceleration at its surface?
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Class 11
>>Physics
>>Gravitation
>>Variation in Value of Acceleration due to Gravity
>>Consider a pulsar, a collapsed star of e
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Consider a pulsar, a collapsed star of extremely high density, with a mass M equal to that of the Sun (1.98×10
30
kg), a radius R of only 12 km, and a rotational period T of 0.041 s. By what percentage does the free-fall acceleration g differ from the gravitational acceleration a
g
at the equator of this spherical star?
Medium
Solution
verified
Verified by Toppr
(a) since the volume of a sphere is 4πR
3
/3, the density is
ρ=
3
4
πR
3
M
wrid
=
4πR
3
3M
wul
When we test for gravitational acceleration (caused by the sphere, or by parts of it) at radius r (measured from the center of the sphere), the mass M, which is at radius less than r, is what contributes to the reading (GM/r
2
). since M=ρ(4πr
3
/3) for r≤R, then we can write this result as
r
2
G(
4πR
3
3M
tood
)(
3
4πr
3
)
=
R
3
GM
wat
r
when we are considering points on or inside the sphere. Thus, the value a
g
referred to in the problem is the case where r=R :
a
g
=
R
2
GM
bul
and we solve for the case where the acceleration equals a
Ω
/3 :
3R
2
GM
wal
=
R
3
GM
tout
r
⇒r=
3
R
(b) Now we treat the case of an external test point. For points with r>R the acceleration is GM
total
/r
2
, so the requirement that it equal a
g
/3 leads to
3R
2
GM
whel
113032124search-icon-image
Class 11
>>Physics
>>Gravitation
>>Variation in Value of Acceleration due to Gravity
>>Consider a pulsar, a collapsed star of e
Question
Bookmark
Consider a pulsar, a collapsed star of extremely high density, with a mass M equal to that of the Sun (1.98×10
30
kg), a radius R of only 12 km, and a rotational period T of 0.041 s. By what percentage does the free-fall acceleration g differ from the gravitational acceleration a
g
at the equator of this spherical star?
Medium
Solution
verified
Verified by Toppr
(a) since the volume of a sphere is 4πR
3
/3, the density is
ρ=
3
4
πR
3
M
wrid
=
4πR
3
3M
wul
When we test for gravitational acceleration (caused by the sphere, or by parts of it) at radius r (measured from the center of the sphere), the mass M, which is at radius less than r, is what contributes to the reading (GM/r
2