(a) If the legendary apple of Newton could be released from rest at a height of 2.5 m from the surface of a neutron star with a mass 1.7 times that of our sun (whose mass is 1.99 x 1030 kg) and a radius of 19 km, what would be the apple's speed when it reached the surface of the star? (b) If the apple could rest on the surface of the star, what would be the difference between the gravitational acceleration at the top and at the bottom of the apple? Take the apple to be a sphere with a radius of 3.6 cm.

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Asked Nov 17, 2019
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(a) If the legendary apple of Newton could be released from rest at a height of 2.5 m from the surface of a neutron star with a mass 1.7 times that of our sun (whose mass is 1.99 x 1030 kg) and a radius of 19 km, what would be the apple's speed when it reached the surface of the star? (b) If the apple could rest on the surface of the star, what would be the difference between the gravitational acceleration at the top and at the bottom of the apple? Take the apple to be a sphere with a radius of 3.6 cm.

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Expert Answer

Step 1

Write the formula for the gravitational potential energy.

-GM
U =
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-GM U =

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Step 2

(a)

Consider r be the distance from the center to the surface and R be the distance from the center to the initial position.  Also, the apple released from the rest. Therefore, the initial kinetic energy is zero.

From conservation of energy.

KE,+U, KE, U
GM 1
GM t
2
1
GM
GM
R
(R-r
2GM,
rR
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KE,+U, KE, U GM 1 GM t 2 1 GM GM R (R-r 2GM, rR

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Step 3

Substitute the known val...

2(6.67x101 N m2/kg? )(1.99x1030 kg)(2.5 m)
(19 km)(19 km2.5 m)
2(6.67x101 N m2 /kg )(1.99 x 103 kg)(2.5 m)
1000 m
(19 km
1000 m
(19 km
2.5 m
1 km
1km
1.356x10 m/s
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2(6.67x101 N m2/kg? )(1.99x1030 kg)(2.5 m) (19 km)(19 km2.5 m) 2(6.67x101 N m2 /kg )(1.99 x 103 kg)(2.5 m) 1000 m (19 km 1000 m (19 km 2.5 m 1 km 1km 1.356x10 m/s

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