Question

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.

Answer 1

Answori- rio 19000 +2.5 r2 = 19000m Us - Gmi m2 r se component Company) our Gmma [ + ] 20 = 66100211429, d'on (monges ino) = 1.14x10 m2 AU : 6-67 no But 10l = IAKET 1.Tuxid² m₂ = 1 m2 V2 V= 1.8x10mis b) Gravitational acceleration g- Giy dg = -2GM ar dr= 7.2cm=0.072m 1991 = 2x6.67X1" x 1.7x199x180x0.072 (9000)3 = 4.13* 100 m/s² -2G1Ndr