calculate the acceleration due to gravity on the surface of a pulsar of mass M=1.98×10^30kg and radius R=12km rotating with time period T=0.041 seconds
Answers
Answer:
(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
=
r
2
GM
wull
⇒r=
3
R
Explanation:
कैलकुलेट एक्सीलरेशन ड्यू टो ग्रेविटी ऑफिस ऑफिस ऑफिस