The most energy-efficient way to send a spacecraft to the moon is to boost its speed while it is in circular orbit about the earth such that its new orbit is an ellipse. the boost point is the perigee of the ellipse, and the point of arrival at the moon is the apogee. calculate the percentage increase in speed required to achieve such an orbit. assume that the spacecraft is initially in a low-flying circular orbit about earth. the distance between earth and the moon is approximately 60 re, where re is the radius of earth

98.33 %

Explanation:

On an elliptical orbit, angular momentum will be conserved .

Angular momentum = I ω = mvR

So mv₁R₁ = mv₂R₂

= v₁R₁ = v₂R₂

where v₁ is velocity and R₁ radius in low orbit (perigee)and v₂ and R₂ is velocity and radius in high orbit ( apogee ).

Here R₁ = Radius of the earth , R₂ is distance between moon and earth.

R₁ / R₂ = 1/60

v₁ /v₂ = R₂ / R₁  = 60

v₂ / v₁ = 1 / 60

1 - (v₂ / v₁ ) = 1 -( 1 / 60)

(v₁ -v₂)/v₁ = ( 60-1 )/60

(v₁ -v₂)/v₁ x 100 = 5900/60 = 98.33 %

total distance travelled = 60 miles

total time taken =1.5 hours

average speed = total distance / total time

60 / 1.5

= 40

therefore, average speed = 40 m/h

happy to

pls mark as brainliest

absorption lines are where light has been absorbed by the atom thus you see a dip in the spectrum whereas.

emission spectra have spikes in the spectra due to atoms releasing photons at those wavelengths.

explanation: