The small spherical planet called \"Glob\" has a mass of 7.96*10^18 kg and a rad

Question

The small spherical planet called "Glob" has a mass of 7.96*10^18 kg and a radius of 6.08*10^4 m. An astronaut on the surface of Glob throws a rock straight up. The rock reaches a maximum height of 1.72*10^3 m, above the surface of the planet, before it falls back down. What was the initial speed of the rock as it left the astronaut's hand? (Glob has no atmosphere, so no energy is lost to air friction. G = 6.67*10-11 Nm2/kg2.) (in m/s)

B.A 36.0 kg satellite is in a circular orbit with a radius of 1.50

Explanation / Answer

conserving energy we get

Ui+Ki= Uf+Kf

-GMm/R+ 1/2*mv^2= -GMm/(R+h)+ 0

thus v^2= 2GM( 1/R - 1/(R+h)

or v^2= 2*GM*h/(r*(R+h))

putting value we get

V^2= 2*7.96*10^18*6.67*10^-11*1.72*10^3/(6.08*10^4*6.252*10^4)

or V=21.9 m/s

B). mv^2/d= GMm/R^2

V^2= GM/R

so V= 6.67*10^-11*7.96*10^18/(1.510^5)

so V=59.56 m/s

Related Questions

Navigate

• Browse (All)
• Browse T
• Subjects

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.